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A TEXTBOOK 



ON 



MECHANICAL DRAWING 

International Correspondence Schools 

SCRANTON, PA. 



ELEMENTARY MECHANICS 

STRENGTH OF MATERIALS 

APPLIED MECHANICS 

MACHINE DESIGN 



SCRANTON 

INTERNATIONAL TEXTBOOK COMPANY 

A-4 



LIBR*^* "* CONGRESS 
Two Cooies Received 

JUN 20 1904 
(\ Cooyrieht Entry 

CLASS Ct, xXc. Na 

K ^ ^ I l=> 
COPY B 



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Copyright, 1897, 1898, 1899, by The Colliery Engineer Company. 
Copyright, 1904, by International Textbook Company. 



Elementary Mechanics : Copyright, 1892, 1893 1894, 1896, 1898, by The Colliery 

Engineer Company. 
Strength of Materials: Copyright, 1894, 1899, by THE Colliery Enginee.. 

PANY. 

Applied Mechanics: Copyright, 1893, 1900, by The COLLIERY ENGINEER COMPANY. 

Machine Design : Copyright, 189o, 1900, by The Colliery Engineer Company. 

Elementary Mechanics, Key : Copyright, 1893, 1894, 1895, 1897, by The Colliery 
Engineer Company. 

Strength of Materials, Key : Copyright, 1894, by The Colliery Engineer Com- 
pany. 

Applied Mechanics, Key : Copyright, 1895, by THE COLLIERY ENGINEER COMPANY. 

Machine Design, Key: Copyright, 1896, 1897, by The COLLIERY ENGINEER COM- 
PANY. 



All rights reserved. 



am 



BURR PRINTING HOUSE, 

FRANKFORT AND JACOB STREETS, 

NEW YORK. 



^ y^/ 



CONTENTS 



Elementary Mechanics ' Page 

Matter and its Properties 297 

Motion and Rest 301 

Force 303 

Dynamics 317 

Falling Bodies 320 

Statics 329 

Simple Machines 335 

Work and Energy 352 

Density and Specific Gravity 358 

Strength of Materials 

Materials Used for Construction ' . 741 

• Stresses and Strains 748 

Factors of Safety 756 

Elementary Graphical Statics 763 

Beams 771 

Columns 798 

Torsion and Shafts . 804 

Ropes 807 

Chains 809 

Applied Mechanics 

Link Mechanisms 812 

Cams 839 

Power Transmission by Belt . . . . ... 852 

Wheels in Trains 872 

Gear-Wheels 898 

Ratchet Wheels/.^ 923 

iii 



iv CONTENTS 

Machine Design Page 

Introductory 1215 

Materials Used in Machine Construction . . . 1219 

Fastenings 1224 

Rotating Pieces 1261 

Bearings 1293 

Tooth Gearing 1314 

Friction Gearing 1330 

Belt Gearing 1332 

Rope Belting 1343 

Chains 1359 

Pipe Flanges 1365 



ELEMENTARY MECHANICS. 



MATTER AND ITS PROPERTIES 



DEFINITIONS. 

S28. Matter is anything that occupies space. It is 
the substance of which all bodies are composed. Matter is 
composed of molecules and atoms. 

829. A molecule is the smallest portion of matter that 
can exist without changing its nature. 

830. An atom is an indivisible portion of matter. 
Atoms unite to form molecules, and a collection of mole- 
cules form a mass or body. 

A drop of water may be divided and subdivided, until 
each particle is so small that it can only be seen by the most 
powerful microscope, but each particle will still be water. 
Now, imagine the division to be carried on still farther until a 
limit is reached beyond which it is impossible to go without 
changing the nature of the particle. The particle of water 
is now so small that, if it be divided again, it will cease to 
be water, and will be something else ; we call this particle a 
molecule. 

If a molecule of water be divided, it will yield two atoms 
of hydrogen gas, and one of oxygen gas. If a molecule of 
sulphuric acid be divided, it will yield two atoms of hydrO' 
gen, one of sulphur, and four of oxygen. 

It has been calculated that the diameter of a molecule is 
larger than y^^^^Voo-oT of ^^ i^^ch, and smaller than goooooooo 
of an inch. 

831. Bodies are composed of collections of molecules. 
Matter exists in three conditions or forms: solid., liquid^ and 
gaseous. 

For notice of the copyright, see page immediately following the title pagOi 



298 ELEMENTARY MECHANICS. 

832. A solid body is one whose molecules change 
their relative positions with great difficulty ; as iron, wood, 
stone^ etc. 

833. A liquid body is one whose molecules tend to 
change their relative positions easily. Liquids readily adapt 
themselves to the vessel which contains them, and their 
upoer surface always tends to become perfectly level. Water, 
mercury, molasses, etc., are liquids. 

834. A gaseous body, or gas, is one whose molecules 
tend to separate from one another; as air, oxygen, hydro- 
gen, etc. 

Gaseous bodies are sometimes called aeriform (air-like) 
bodies. They are divided into two classes — the so-called 
''^ permanent " gases^ and vapors. 

835. A permanent gas is one which remains a gas 
at ordinary temperatures and pressures. 

836. A vapor is a body which at ordinary tempera- 
tures is a liquid or solid, but, when heat is applied, becomes 
a gas, as steam. 

One body may be in all three states ; as, for example, mer- 
cury, which at ordinary temperatures is a liquid, becomes a 
solid (freezes) at 40° below zero, and a vapor (gas) at 600° 
above zero. By means of great cold, all gases, even hydro- 
gen, have been liquefied, and some solidified. 

By means of heat, all solids have been liquefied and a great 
many vaporized. It is probable that if we had the means of 
producing sufficiently great extremes of heat and cold, all 
solids might be converted into gases, and all gases into solids. 

837. Every portion of matter possesses certain qualities 
Q,2^}i^di properties. Properties of matter are divided into two 
classes : general and special. 

838. General properties of matter are those which 
are common to all bodies. They are as follows : Extension^ 
impenetrability ^ weighty indestructibility ^ inertia^ mobility^ 
divisibility^ porosity^ compressibility^ expansibility ^ and 
elasticity. 



ELEMENTARY MECHANICS. 299 

839. Special properties are those which are not pos- 
sessed by all bodies. Some of the most important are as 
follows: Hardness^ tenacity^ brittleness^ malleability ^ and 
ductility. 

840. Extension is the property of occupying space. 
Since all bodies must occupy space, it follows that extension 
is a general property. 

841. By impenetrability we mean that no two bodies 
can occupy exactly the same space at the same time. 

842. "Weight is the measure of the earth's attraction 
upon a body. All bodies have weight. In former times it 
was supposed that gases had no weight, since, if unconfined, 
they tend to move away from the earth; but, nevertheless, 
they will finally reach a point beyond which they cannot go, 
being held in suspension by the earth's attraction. Weight 
is measured by comparing it with a standard. The stand- 
ard is a bar of platinum- owned and kept by the government; 
it weighs one pound. 

843. Inertia means that a body cannot put itself in 
motion nor bring itself to rest. To do so, it must be acted 
upon by some force. 

844. Mobility means that a body can be changed in 

position by some force acting upon it. 

845. Divisibility is that property of matter which 
indicates that a body may be separated into parts. 

846. Porosity is that property of matter which indi- 
cates that there is space between the molecules of a body. 
Molecules of bodies are supposed to be spherical, and, hence, 
there is space between them, as there would be between 
peaches in a basket. The molecules of water are larger than 
those of salt; so that when salt is dissolved in water, its 
molecules wedge themselves between the molecules of the 
water, and, unless too much salt is added, the water will oc- 
cupy no more space than it did before. This does not prove 
that water is penetrable, for the molecules of salt occupy 
the space that the molecules of water did not. 



300 ELEMENTARY MECHANICS. 

Water has been forced through iron by pressure, thus 
proving that iron is porous. 

847. Compressibility is that property of matter 
which indicates that the molecules of a body may be 
crowded nearer together, so as to occupy a smaller space. 

848. Expansibility is that property of matter which 
indicates that the molecules of a body may be forced apart, 
so as to occupy a greater space. 

849. Elasticity is that property of matter which indi- 
cates that if a body be distorted within certain limits, it will 
resume its original form when the distorting force is re- 
moved. Glass, ivory, and steel are very elastic. 

850. Indestructibility indicates that matter can 
never be destroyed. A body may undergo thousands of 
changes ; be resolved into its molecules, and its molecules 
into atoms, which may unite with other atoms to form other 
molecules and bodies, which may be entirely different from 
the original body, but the same number of atoms remains. 
The whole number of atoms in the universe is exactly the 
same now as it was millions of years ago, and will always be 
the same. Matter is indestructible. 

851. Hardness is that property of matter "Vhich indi- 
cates that some bodies may scratch other bodies. Fluids 
and gases do not possess hardness. The diamond is the 
hardest of all substances. 

852. Tenacity is that property of matter which indi- 
cates that some bodies resist a force tending to pull them 
apart. Steel is very tenacious. 

853. Brittleness is that property of matter which 
indicates that some bodies are easily broken; as glass, 
crockery, etc. 

854. Malleability is that property of matter which 
indicates that some bodies may be hammered or rolled into 
sheets. Gold is the most malleable of all substances. 



ELEMENTARY MECHANICS. 301 

855. Ductility is that property of matter which indi- 
cates that some bodies may be drawn into wire. Platinum 
is the most ductile of substances. 

856. Mechanics is that science which treats of the 
action of forces upon bodies, and the effects which they pro- 
duce ; it treats of the laws which govern the movement and 
equilibrium of bodies, and shows how they may be utilized. 



MOTION AND REST. 



VELOCITY. 

857. Motion is the opposite of rest, and indicates a 
changing of position in relation to some object. If a large 
stone is rolled down hill, it is in motion in relation to the hill. 

If a person is on a railway train, and walks in the opposite 
direction from that in which the train is moving, and with 
the same speed, he will be in motion as regards the train, 
but at rest with respect to the earth, since, until he gets to 
the end of the train, he will be directly over the spot at 
which he was when he started to walk. 

858. The path of a body in motion is the line described 
by its central point. No matter how irregular the shape of 
the body may be, nor how many turns and twists it may 
make; the line which indicates the direction of the center of 
the body for every instant that it was in motion, is the path 
of the body. , 

859. Velocity is rate of motion. It is measured by a 
unit of space passed over in a unit of time. When equal 
spaces are passed over in equal times, the velocity is said to 
be uniform. In all other cases it is variable. 

If the fly-wheel of an engine keeps up a constant speed of 
a certain number of revolutions per minute, the velocity of 
any point on the wheel is uniform. A railway train having 
a constant speed of 40 miles per hour, moves 40 miles every 
hour, or l-g^ = f of a mile every minute ; and, since equal 
spaces are passed over in equal times, the velocity is uniform, 



302 ELEMENTARY MECHANICS. 

Let S = the length of space passed over uniformly; 

/ = the time occupied in passing over the space S; 
V= the velocity. 
Then, the velocity V must equal the space 5", divided by 
the time /, or 

^=T (^•) 

Also, the space 5 must eqval the velocity F, multiplied by 
the time, or 

S=Vf. (8.) 

The time ^ must equal the space 5, divided by the veloc- 
ity, or 

t=y (9.) 

860. Unless stated otherwise, the space passed over will 
be the length of the path of the body^ and will be measured 
in feet and decimals of a foot, and, unless otherwise stated, 
the time will be measured in seconds. 

When these units are used, the velocitywill be in feet per 
second^ which means that the center of the body passed over 
a certain number of feet every second^ and the unit will be 
one foot in one second. 

Example. — The velocity of sound in still air is 1,092 feet per second. 
If I see the flash of a cannon when it is fired, but do not hear the report 
until 5 seconds afterwards, how far away is the cannon ? 

Solution.— 6" =Vt — 1,092 x 5 = 5,460 feet. Ans. 

ExAJiPLE. — The velocity of light is 186,000 miles a second. If the 
average distance from the earth to the sun is 93,(500,000 miles, how long 
does it take for a beam of light to reach the earth from the sun ? 

„ , S 93,000,000 ^^„ - Q • ^ on 

Solution. — t = -y^ = '^ ' — 500 seconds, or 8 mmutes 20 
V 186,000 

seconds. Ans. 

Example. — If a body passes over a space of 4,800 feet uniformly in 8 
minutes, what is its velocity in feet per second ? 

Solution. — 8 minutes = 480 seconds. V — — — ' = 10. Hence, 

/ 4oO 

the velocity is 10 feet per second. Ans. 

In examples concerning work the unit of velocity is usually 
taken as one foot in one minute. 



ELEMENTARY MECHANICS. 303 

The unit of time may be a second^ minute^ hour^ day, or 
year. The unit of space may be feet.^ miles, the earth's 
radius, or the distance from the earth to tJie sun, according 
to the conditions of the example. The larger units are used 
only in astronomy. 

Example. — The distance from the earth to the moon is about 60 
times the radius of the earth ; how many miles is it from here to the 
moon ? 

Solution. — The radius of the earth is nearly 4,000 miles; hencQ^ 
4,000 X 60 = 240,000 miles, the distance to the moon, nearly. Ans. 



EXAMPLES FOR PRACTICE. 

1. The piston speed of a steam engine is 10 feet per second ; how 
many miles will the piston travel in one hour ? Ans. 6j\ mi. 

2. If a railroad train travels 70 miles in one hour, what is its velocity 
in feet per second ? Ans. 102f ft. per sec. 

3. A man runs 100 yards in 12 seconds ; how long will it take him 
to run a mile at the same rate ? Ans. 3 min. 31.2 sec. 

4. The outside diameter of an engine fly-wheel is 13 feet 9 inches. 
A point on the rim travels 45,000 feet in 5 minutes; what is the velocity 
in feet per second ? Ans. 150 ft. per sec 

FORCE. 



THE THREE LAWS OF MOTION. 

S61. A force is that which produces, or tends to pro- 
duce or destroy, motion. Forces are called by various 
names, according to the effects which they produce upon a 
body, as attraction, ? epulsion, cohesion, adhesion, accelerating 
force, retarding force, resisting force, etc., but all are 
equivalent to a push or pull, according to the direction in 
which they act upon a body. That the effect of a force 
upon a body may be compared with another force, it is 
necessary that three conditions be fulfilled in regard to both 
forces; they are as follows: 

(1.) The point of application^ or point at which the force 
acts upon the body, must be known. 

(2.) The direction of the force, or, what is the same thing, 
the straight line along which the force tends to move the 
point of application, must he known. 



du4 ELEMENTARY MECHANICS. 

(3.) TJie magnitiLde or value of tJic force^ when cant' 
pared zvitJi a given standard^ must be knozvn. 

862. The unit of magnitude of forces will always be 
taken as one pound ^ in this section on Elementary Mechanics, 
and all forces will be spoken of as a certain number of pounds. 

863. According to the effects which forces produce upon 
a body, the science of Mechanics is subdivided as follows: 

( 1 . ) Median ics of Solid Bodies. 

(2.) Mechanics of Fluids. 

(3.) Mechanics of Heat ^ or Thermodynamics. 

Mechanics of Solids is further divided into Statics and 
Kinetics^ or Dynamics^ as it is commonly called. 

Mechanics of Fluids is further divided into Mechanics oj 
Air and Gases ^ or Pneumatics^ and Mechanics of Liquids. 
The Mechanics of Liquids is divided into Hydrostatics and 
Hydrokinetics ; the latter is also called Hydraulics and 
Hydrodynam ics. 

864. Statics treats of the conditions of the equilibrium 
of bodies. A body is in equilibrium under the action of 
forces, when the forces acting upon the body balance each 
other. 

865. Kinetics, or Dynamics, treats of bodies in 
motion, and the effects which they may produce. 

866. Pneumatics treats of the laws of the pressure 
and of the movement of air and other gaseous bodies. 

867. Hydrostatics treats of the equilibrium of liquids. 

868. Hydrokinetics (also called Hydraulics and 
Hydrodynamics) treats of liquids in motion, and the effects 
which they may produce. 

869. Thermodynamics treats of the mechanical 
effects of heat upon bodies. 

870. The fundamental principles of the relations 
between force and motion were first stated by Sir Isaac 
Newton. They are called " Newton's Three Laws of 
Motion," and are as follows: 



ELEMENTARY MECHANICS. 305 

(I.) All bodies continue in a state of rcst^ or of uniform 
motion^ in a straight line^ unless acted upon by some external 
force that compels a change. 

(11. ) Every motion^ or change of motion^ is proportio7ial to 
the acting force ^ and takes place in the direction of the 
straight line along which the force acts. 

(III.) To every action there is always opposed an equal 
and contrary reaction. 

From the first law of motion^ it is inferred that a body 
once set in motion by any force, no matter how small, will 
move forever in a straight line, and always with the same 
velocity, unless acted upon by some other force which com- 
pels a change. It is not possible to actually verify this law, 
on account of the earth's attraction for all bodies, but, from 
astronomical observations, we are certain that the law is 
true. This law is often called tJie law of inertia. 

871 . The word inertia is so abused that a full under- 
standing of its meaning is necessary. Inertia is not a force, 
although it is often so called. If a force acts upon a body 
and puts it in motion, the effect of the force is stored in the 
body ; and a second body, in stopping the first, will receive a 
blow equal in every respect to the original force, assuming 
that there has been no resistance of any kind to the motion 
of the first body. 

It is dangerous for a person to jump from a fast-moving 
train, for the reason that, since his body has the same 
velocity as the train, it has the same force stored in it that 
would cause a body of the same weight to take the same 
velocity as the train, and the effect of a sudden stoppage is 
the same as the effect of a blow necessary to give the person 
that velocity. But, by "bracing " himself and jumping in 
the same direction that the train is moving, and running, 
he brings himself gradually to rest, and thus reduces the 
danger. If a body is at rest, it must be acted upon by a 
force in order to be put in motion, and, no matter how 
great the force may be, the body cannot be instantly put in 
motion. 



306 



ELEMENTARY MECHANICS. 



The resistance thus offered to being put in motion is com. 
monly, but erroneously, called, the Resistance of Inertia. 
It should be called the Resistance due to Inertia. 

From the second law^ it is seen that, if two or more forces 
act upon a body, their final effect upon the body will be in 
proportion to their magnitude and to the directions in which 
they act. Thus, if the wind is blowing due west, with a 
velocity of 50 miles per hour, and a ball is thrown due north 

with the same velocity, 



or 50 miles per hour, 
the wind will carry the 
ball just as far west as 
the force of the throw 
carried it north, and 
the combined effect will 
be to cause it to move 
north-west. The amount 
of departure from due 
north will be propor- 
tional to the force of 
the wind, and independ- 
ent of the velocity due 
to the force of the 
throw. 

In Fig. 110,. a ball e 
is supported in a cup, 
the bottom of which is 
attached to the lever a 
in such a manner that 
a movement of o will 
swing the bottom hori- 
zontally and allow the 
ball to drop. Another 
ball b rests in a hori- 
zontal groove that is 
provided with a slit in 
the bottom. A swing- 
ing arm is actuated by 




ELEMENTARY MECHANICS. 307 

the spring d in such a manner that, when drawn back as 
shown and then released, it will strike the lever o and the 
ball b at the same time. This gives b an impulse in a hor- 
izontal direction and swings o so as to allow e to fall. 

On trying the experiment, it is found that b follows a 
path shown by the curved dotted line, and reaches the floor 
at the same instant as ^, which drops vertically. This 
shows that the force which gave the first ball its horizontal 
movement, had no effect on the vertical force which com- 
pelled both balls to fall to the floor, the vertical force produc- 
ing the same effect as if the horizontal force had not acted. 
The second law may also be stated as follows: A force has 
the same effect in producing motion^ whether it acts upon a 
body at rest^ or in motion^ and whether it acts alone or with 
other forces. 

The third law states that action and reaction are equal 
and opposite. A man cannot lift himself by his boot-straps, 
for the reason that he presses downwards with the same 
force that he pulls upwards ; the downward reaction equals 
the upward action, and is opposite to it. 

In springing from a boat we must exercise caution, or the 
reaction will drive the boat from the shore. When we jump 
from the ground, we tend to push the earth from us, while 
the earth reacts and pushes us from it. 

872. A force may be represented by a line ; thus, in 
Fig. Ill, let A be the point of application of the force; let 
the length of the line A B represent its 
magnitude^ and let the arrow-head indicate * ^ 

the direction in which the force acts, then fig. in. 

the line A B fulfils the three conditions (see Art. 861), and 
the force is fully represented. 



THE COMPOSITION OF FORCES. 

873. When two forces act upon a body at the same 
time, but at different angles, their final result may be ob- 
tained as follows: 



308 



ELEMENTARY MECHANICS. 



50 Ih. 



B 




In Fig. 112, let A be the common point of application of 
the two forces, and let A B and A C represent the magni- 
tude and direction of the forces. Ac- 
cording to the second law of motion, 
the final effect of the movement due 
to these two forces would be the same 
whether they acted singly or together. 
Suppose that the line A B represents 
the distance that the force A B would 
cause the body to move; similarly, 
that A C represents the distance 
Fig. 112. which the force A C would cause the 

body to move when both forces were acting separately. 
The force A B^ acting alone, would carry the body to ^; if 
the force A C were now to act upon the body, it would 
carry it along the line B D, parallel to A C, to a, point D^ 
at a distance from B equal to A C. Join C and D^ then 
C D v^ parallel to A B^ and A B D C \s> a. parallelogram. 
Draw the diagonal A D. According to the second law of 
motion, the body will stop at D^ whether the forces act 
separately or together, but if they act together, the path 
of the body will be along A D, the diagonal of the paral- 
lelogram. Moreover, the length of the line A D repre- 
sents the magnitude of a force which, acting at A in the 
direction A D, would cause the body to move from A to D\ 
in other words, A D^ measured to the same scale as ^ ^ and 
A (7, represents, in magnitude and direction^ the combined 
effect of the two forces A B and A C. 



874. This line A D \'s> called the resultant. Suppose 

that the scale used was 50 pounds to the inch ; then, \i A B 

= 50 pounds, and A C = 62|- pounds, the length oi A B 

50 62 5 

would be — =1 inch, and the length oi A C would be -—-- 
50 50 

= 1:^ inches. If A D, or the resultant^ measures If inches, 

its magnitude would be If x 50 = 87|- pounds. 

Therefore, a force of 87|- pounds acting upon a body at A 
in the direction A D^ will produce the same result as the 



ELEMENTARY MECHANICS. 309 

combined effects of a force of 50 pounds acting in the direc- 
tion A B, and a force of 62|- pounds acting in the direction A C. 

875. The above method of finding the resuhing action 
of two forces acting upon a body at a common point, is correct, 
whatever may be their direction and magnitudes. Hence, 
to find the resultant of two forces when their common point 
of appUcation, their direction and magnitudes are known : 

Rule I. — Assume a pointy and draw two lines parallel to 
the directions of the lines of action of the two forces. With 
any convenient scale ^ measure off from t lie point of intersec- 
tion (common point of application^^ distances corresponding to 
the magnitudes of the respective forces^ and co^nplete the 
parallelogram. From tJie common point of application^ draw 
the diagonal of the parallelogram ; this diagonal will be the 
resultant^ and its direction will be away from the point of 
application. Its magnitude should be measured with the same 
scale that was used to measure the two forces. 

This method is called the graphical metliod of the 
parallelogram of forces. 

876-. Experimental Proof. — The principle of the par- 
allelogram of forces is clearly shown in Fig. 113. A B D C 
is a wooden frame, jointed to allow motion at its four corners. 
The length A B equals C D\ that oi A C equals B D^ and 
the corresponding adjacent sides are in the ratio of two to 
three. Cords pass over the pulleys M and iV, carrying 
weights W Sind w, of 90 and 60 pounds. The ratio between 
the weights equals the ratio of the corresponding adjacent 
sides. A weight V of 120 pounds is hung from the corner A, 

When the frame comes to rest, the sides A B and A C lie 
in the direction of the cords. These sides A B and A C are 
accurate graphic representations of the two forces acting 
upon the point A. It will be found that the diagonal A D 
is vertical, and twice as long 2^^ A C \ hence, since A C rep- 
resents a force of 60 pounds, A D will represent a force of 
2 X 60, or 120 pounds. 

Thus, we see that the line A D represents the resultant of 
r.he two forces A B and A C ; in other words, it represents 
D. 0. ni.—2 



r 



310 



ELEMENTARY MECHANICS. 



the resultant of the two weights W send w. This resultant 
is equal and opposite to the vertical force, which is due to 
the weight of F. 




Fig. 113. 

Satisfactory results of this kind will be secured when we 
have the proportion 

A B'.A C=W \w. 

Example. — If two forces act upon a body at a common point, both 
acting away from the body, and the angle between them is 80°, what is 
the value of the resultant, the magnitude of the two forces being 
60 pounds and 90 pounds, respectively ? 

Solution. — Draw two indefinite lines, Fig. 114, making an angle 
of 80°. With any convenient scale, say 10 pounds to the inch, measure 
off ^ i? = 60 -^ 10 = 6 inches, and ^ C = 90 -v- 10 = 9 inches. 

Through B, draw B D parallel to A C, and through C, draw CD 
parallel to A B, intersecting at B. Then draw A D, and A D will be the 
resultant; its directioti is towards the point D^ as shown by the arrow. 

Measuring A Dy we find that its length = 11.7 inches. Hence, 
11.7 X 10 = 117 pounds. Ans. 

Caution. — In solving problems by the graphical method, use as 
large a scale as possible. More accurate results are then obtained. 

S17.m The above example might also have been solved 




ELEMENTARY MECHANICS. 3H 

by the method called the triangle of forces, which is 
as follows: 

In Fig. 114, suppose that the two forces acted separately, 
first from A to B^ and then 
from B to D^ in the direction 
of the arrows. 

Draw A D\ then ^ Z> is 
the resultant of the forces 
A B and A C, since B D — 
A C; but ^ Z^ is a side of 
the triangle A B D. It will ^ 
also be noticed that the di- 
rection oi A Dv=> opposed to that oi A B and B D ; hence, to 
find the resultant of two forces acting upon a body at a 
common point, by the method of triangle of forces : 

Rule II. — Draw the lines of action of the two forces as 
if each force acted separately^ the lengths of the lines beifig 
proportional to the magnitude of the forces, foin the extrem- 
ities of the two lines by a straight line^ and it will be the re- 
sultajit ; its direction will be opposite to that of the two forces. 

Note. — When we speak of the resultant being opposed in direction 
to the other forces around the polygon, we mean that, starting from 
the point where we began to draw the polygon, and tracing each line 
in succession, the pencil will have the same general direction around 
the polygon, as if passing around a circle, from left to right, or from 
right to left, but that the closing line or resultant must have an oppo- 
site direction, that is, the two arrow-heads must point towards the 
point of intersection of the resultant and the last side. 

878. When three or more forces act upon a body at a 
given point, their resultant may be found by the following rule : 

Rule III. — Find the resultant of any two forces ; treat 
this resultant as a single force ^ and combine it with a third 
force to find a second resultant. Combine this second result- 
ant with a fourth force, to find a third resultant, etc. After 
all the forces have been thus combined, the last resultant will 
be the resultant of all of the forces, both in magnitude and 
direction. 

Example. — Find the resultant of all the forces acting on the point O 
in Fig. 115, the length of the lines being proportional to the magnitude 
of the forces. 



312 



ELEMENTARY MECHANICS. 



Solution. — Draw O E parallel and equal to A O, and EF parallel 
and equal to BO, then O Eis the resultant of these two forces, and its 
direction is from O to E, opposed to O E and E E. Treat O EslsU O E 
and EE did not exist, and draw EG parallel and equal to O C\ O G 
will be the resultant of 0/^and EG\ but O E 'v& the resultant oi O E 
and E E, hence, O G \^ the resultant of O E, EE, and EG, and likewise 
of A O, B O, and CO. The line EG, parallel to CO, could not be drawn 
from the point O to the right of O E, for in that case it would be opposed 
in direction to OE; but EG must have the same direction as O E, in 
order that the resultant may be opposed to both O Eand EG. 

For the same reason, draw GL parallel and equal to EO. Join O 
and L, and O L will be the resultant of all the forces AO, BO, CO., 
and DO (both in magnitude and direction), acting at the point O, If 




L O were drawn parallel and equal to O L, and having the same direc- 
tion, it would represent the effect produced on the body by the com- 
bined action of the forces AO, BO, CO, and DO. 

879. In the last figure, it will be noticed that O E, E F, 
F G^ G L, and L O are sides of a polygon O E F G L/\n 
which O L, the resultant, is the closing side, and that its 
direction is opposed to that of all the other sides. This fact 
is made use of in what is called the metliod of the poly- 
gon of forces. 

To find the resultant of several forces acting upon a body 
at the same point : 

Rule IV. — Through a convenient point on the drawing, 
draw a line parallel to one of the forces^ and having the same 



ELEMENTARY MECHANICS. 



313 



direction and magnitude . At the end of this line, draw an- 
other line parallel to a second force, and having the same di- 
rection and magnitude as this second force ; at the end of the 
second line, draw a line parallel and equal in magnitude and 
direction to a third force. Thus continue until lines have 
been drawn parallel and equal in magnitude and direction to 
all of the forces. 

The straight line joining the free ends of the first and last 
lines will be the closing sides of tJie polygon ; mark it opposite 
in direction to that of the other forces around the polygon^ 
and it will be. the resultant of all the forces. 

Example. — If five forces act upon a body at angles of 60°, 120°, 180°, 
240°, and 270°, towards the same point, and their respective magnitudes 
are 60, 40, 30, 25, and 20 pounds, find the magnitude and direction of 
their resultant by the method of polygon of forces.* 

Solution. — From a common point O, Fig. 116, draw the lines of 





Pig. 116. 

action of the forces, making the given angles with a horizontal line 
through O, and mark them as acting towards O, by means of arrow- 
heads, as shown. Now, choose some convenient scale, such that the 
whole figure may be drawn in a space of the required size on the 
drawing. Choose any one of the forces, as A 0, and draw O F parallel 
to it, and equal in length to 30 pounds on the scale. It must also act in 
the same direction as 6>^- At /^ draw FG parallel to BO, and equal 
to 40 pounds. In a similar manner, draw G H, HI, and /^parallel to 

*NoTE. — As stated in Art. 742, all angles are measured from a 
horizontal line, in a direction opposite to the movement of the hands of 
a watch (from around the circle to the left), from 1° or less up to 360°. 



314 ELEMENTARY MECHANICS. 

CO, DO, and EO, and equal to 60, 20, and 25 pounds, respectively. 
Join O and K hy O K, and O K will be the resultant of the combined 
action of the five forces ; its direction is opposite to that of the other 
forces around the polygon O F G H I K, and its magnitude = 55| 
pounds. Ans. 

If the resultant O K, in Fig. 116, were to act alone upon 
the body in the direction shown by the arrow-head, with a 
force of 55f pounds, it would produce exactly the same 
effect upon a body as the combined action of the five forces. 

li O F, F G^ G //, // /, and / K represent the distances 
and directions that the forces would move the body, if act^ 
ing separately, O K i?, the direction and distance of move^ 
ment of the body when all the forces act together. 

880. From what has been said before, it is seen that 
any number of forces acting on a body at the same point, 
or having their lines of action pass through the same point, 
can be replaced by di single force (resultant), whose line of 
action shall pass through that point. 

881. Heretofore, it has been assumed that the forces 
acted upon a single point on the surface of the body, but it 
will make no difference where they act, so long as the lines 
of action of all the forces intersect at a single pointy either 
within or without the body, only so that the resultant can 
be drawn through the point of intersection. If two forces 
act upon a body in the same straight line and in the same 
direction, their resultant is the sum of the two forces ; but, if 
they act in opposite directions, their resultant is the differ- 
ence of the two forces^ and its direction is the same as that 
of the greater force. If they are equal and opposite, the 
resultant is zero^ or one force just balances the other. 

Example. — Find the resultant of the forces whose lines of action 
pass through a single point, as shown in Fig. 117. 

Solution. — Take any convenient pointy, and draw a line ^y, par- 
allel to one of the forces, say the one marked 40, making it equal in 
length to 40 pounds on the scale, and indicate its direction by the 
arrow-head. Take some other force — the one marked 37 will be con- 
venient; the XwiQfe represents this force. From the point e, draw a 
line parallel to some other force ; say the one marked 29, and make it 
equal in magnitude and direction to it. So continue with the other 
forces, taking cai e that the general direction around the polygon is not 



ELEMENTARY MECHANICS. 



315 



changed. The last force drawn in the figure is a b, representing the 
force marked 25. Join the points a and g\ then, ag is the resultant 
of all the forces shown in the figure. Its direction is from g to a op- 




FiG. 117. 



posed to the general direction of the others around the polygon. It does 
not matter in what order the different forces are taken, the resultant will 
be the same in magnitude and direction, if the work is done correctly. 

882. The various methods of finding the resultant of 
several forces are all grouped under one head : The compo- 
sition of forces. 



THE RESOLUTION OF FORCES. 

883. Since two forces can be combined to form a single 
resultant force, we may also treat a single force as if it were 
the resultant of two forces, whose action upon a body will 
be the same as that of a single force. Thus, in Fig. 118, 
the force O A maybe resolved into two forces, O B a.ndBAy 
whose directions are opposed to O A. 

If the force O A acts upon a body, moving or at rest upon 
a horizontal plane, and the resolved force O B is vertical, 
and B A horizontal, O B^ measured to the same scale as 




316 ELEMENTARY MECHANICS. 

O A^ IS, the magnitude of that part oi O A which pushes the 

body downwards^ while B A is the magnitude of that part 
Q of the force O A which is 

exerted in pushing the body 
in a horizontal direction. 
O B and B A are called the 
components of the force 
O A^ and when these com- 
ponents are vertical and 
fi'g. 118. horizontal, as in the present 

case, they are called the vertical component and the horizontal 

component of the force O A. 

884. It frequently happens that the position, magni- 
tude and direction of a certain force is known, and that it is 
desired to know the effect of the force in some direction, 
other than that in which it acts. Thus, in Fig. 118, suppose 
that O A represents, to some scale, the magnitude, direction, 
and line of action of a force acting upon a body at ^, and 
that it is desired to know what effect O A produces in the 
direction B A. Now B A, instead of being horizontal, as in 
the cut, may have any direction. To find the value of the 
component oi O A which acts in the direction B Ay wq 
employ the following rule : 

Rule V. — From one extreinity of the line representing the 
given force ^ draw a line parallel to tJie direction in which it is 
desired that the component shall act ; from the other extrem- 
ity of the given force ^ draw a line perpendicular to the com^ 
ponent first drawn^ and intersecting it. The length of the 
component y measured from the point of intersection to the in- 
tersection of the component with tJie given force ^ will be mag- 
nitude of the effect produced by the given force in the required 
direction. 

Thus, suppose O A, Fig. 118, represents a force acting 
upon a body resting upon a horizontal plane, and it is de- 
sired to know what vertical pressure O A produces on the 
body. Here the desired direction is vertical; hence, from 
one extremity, as (9, draw O B parallel to the desired direc- 
tion (vertical in this case), and from the other extremity, 



ELEMENTARY MECHANICS. 



317 



draw A B perpendicular to O B^ and intersecting B 2X 
B. Then O B^ when measured to the same scale as (9 ^, 
will be the value of the vertical pressure produced by O A. 

Example. — If a body weighing 200 pounds rests upon an inclined 
plane whose angle of inclination to the horizontal is 18°, what force 
does it exert perpendicular to the plane, and what force does it exert 
parallel to the plane, tending to sUde downwards ? 

SoLUTiON. — Let ABC, Fig. 119, be the plane, the angle A being 
equal to 18", and let W be the weight. Draw a vertical line FD =a 
200 pounds, to represent the magnitude 
of the weight. Through 7% draw FE 
parallel to A By and through D draw 
/? ^ perpendicular to E F^ the two lines 
intersecting at E. F D is now resolved 
into two components, one, FE, tending 
to pull the weight downwards, and the 
other, ED, acting as a perpendicular 
pressure on the plane. 

Since FB is perpendicular to A C, 
aud ^Z> is perpendicular to A B^ the 
angle D = angle A - 18°. 

Hence, FE- 200 X sin 18° = 200 X .30903 = 61.804 pounds, and ED 
c: 200 X cos 18° = 200 X .95106 = 190.212 pounds. 

Force parallel to the plane = 61.804 pounds. \ . 

Force perpendicular to the plane = 190.212 pounds. \ 




DYNAMICS. 

885. Dynamics may be defined as that branch of 
Mechanics which deals with bodies moving with a variable 
velocity. In Elementary Mechanics we shall consider only 
falling bodies and centrifugal force. 



GRAVITATION. 

886. Every body in the universe exerts a certain 
attractive force on every other body, which tends to draw 
the two bodies together. This attractive force is called 
gravitation. 

If a body is held in the hand, a downward pull is felt, and 
if let go of, it will fall to the ground. This pull is com- 
monly called weighty but it really is the attraction between 
the earth and the bod v. 



318 ELEMENTARY MECHANICS. 

887. Force of gravity is a term used to denote the 
attraction between the earth and bodies upon or near its 
surface. It always acts in a straight line between the cen- 
ter of the body and the center of the earth. The force of 
gravity varies at points on the earth's surface. 

It is slightly less on the top of a high mountain than at 
the level of the sea. For this reason, the weight of a body 
also varies. But if the weight of a body at any place be 
divided by the force of gravity at that place, the result is 
called the mass of the body. 

888. The mass of a body is the measure of the actual 
amount of matter that it contains, and is always the same. 

If the mass of the body be represented by m, its weight 

by W^ and the force of gravity at the place where the body 

was weighed, by g^ we have 

weiofht of body W ,^ ^ ^ 

mass = 7 ^ — 7 rr^, or^ = — . (10.) 

force of gravity g ^ ' 

889. La^w of Gravitation : — 

The force of attraction by which one body tends to draw 
another body towards it^ is directly proportional to its mass^ 
and inversely proportional to the square of the distance be- 
tween their centers. 

890. Laws of Weight : — 

Bodies weigh most at the surface of the earth. Below the 
surface^ the weight decreases as the distance to the center 
decreases. 

Above the surface the weight decreases as the square of the 
distance increases. 

Illustration. — If the earth's radius is 4,000 miles, a body 
that weighs 100 pounds at the surface will weigh nothing at 
the center, since it is attracted in every direction with equal 
force. At 1,000 miles from the center, it will weigh 25 
pounds, since 

4,000 : 1,000 = 100 : 25. 
At 2,000 miles from the center, it will weigh 50 pounds, 
since 

4,000 : 2,000 = 100 : 50. 



ELEMENTARY MECHANICS. 319 

At 3,000 miles from the center, it will weigh 75 pounds, 
and at the surface, or 4,000 miles from the center, it will 
weigh 100 pounds. If carried still higher, say 1,000 miles 
from the surface, or 5,000 miles from the center of the 
earth, it will weigh 64 pounds, since 

5,000' : 4,000' = 100 : 64. 

At 4,000 miles from the surface, it will weigh 25 pounds, 

since 

8,000' : 4,000' = 100 : 25. 

891. Formulas for Gravity Problems : — 

Let W= weight of body at the surface; 

w = weight of a body at a given distance above or 

below the surface; 
d= distance between the center of the earth and 

the center of the body ; 
R = radius of the earth = 4,000 miles. 
Formula for weight when the body is below the surface: 

wR = dlV. (11.) 

Formula for weight when the body is above the surface : 
wd'=lVR\ (12.) 

Example. — How far below the surface of the earth will a 26-pound 
ball weigh 9 pounds ? 

Solution. — Use formula 11, wR = d IV, 
Substituting the values of 7?, W^ and w, we have 
9x4,000 = ^X25, or 

</= 5^ = 1,440 miles from the center. Ans. 

Example. — If a body weighs 700 pounds at the surface of the earth, 
at what distance above the earth's surface will it weigh 112 pounds ? 

Solution. — Use formula 12, ^^'^= ^i?^ 
Substituting the values of R, IV, and w, we have 
11 2 Xd^ = 700 X 4,0002, or 

^=|/M^|M = 10.000 miles. 

Therefore, 10,000 — 4,000 = 6,000 miles above the earth's surface. 
Ans. 

Example. — The top of Mt. Hercules was said to be 32,000 feet, say 
6 miles above the level of the sea. If a body weighs 1,000 pounds at 
sea-level, what would it weigh if carried to the top of the mountain ? 

Solution.— wi/*= IV R\ or «/ x 4,006^ = 1,000 x 4,000« ; whence, 

4,000»X 1,000 onrr ^ A 

» = — — A /v^/>. = ®^^ pounds. Ans. 

4.006' 



320 



ELEMENTARY MECHANICS. 



EXAMPLES FOR PRACTICE. 

1. How much would 1,000 tons of coal weigh one mile below 
the surface? Ans. 1,999,5001b. 

2. How much would the coal in example 1 weigh one mile above the 
surface? Ans. 1,999,000 lb., nearly. 

3. How far above the earth's surface would it be necessary to cairy 
a body in order that it may weigh only half as much ? 

Ans. 1,656.854 miles, nearly. 

4. A man weighs 160 pounds at the surface; how much will he 
weigh 50 miles below the surface ? Ans. 158 lb. 

5. If a body weighs 100 pounds 400 miles above the earth's surface, 
how much will it weigh at the surface ? Ans. 121 lb. 

Note. — Use 4^000 mileiixs the radius of the earth. 



FALLING BODIES. 

892. If a leaden ball and a piece of paper are dropped 
from the same height, the ball would strike the ground first. 

This is not because the leaden ball is the 
heavier, but because the resistance of the air 
has a greater retarding effect upon the paper 
than upon the ball. If we place this same 
leaden ball and a piece of paper in a glass 
tube, Fig. 120, from which all of the air has 
been exhausted, it would be found that when 
the tube was inverted, both would drop to 
the bottom in exactly the same time. This 
experiment proves that it was only the resist- 
ance of the air that caused the ball to reach 
the ground first, in the former experiment. 
This resistance of the air may be nearly 
equalized by making the two bodies of the 
same shape and size. For example, if a 
wooden and an iron ball, having equal diam- 
eters, were dropped from the same height, 
they would strike the ground at almost 

exactly the same instant, although the iron ball might be 

ten times as heavy as the wooden ball. 

893. Suppose there were several leaden balls, as shown 
in Fig. 121, at ^; it is obvious that if they were dropped 




Fig. 120. 



ELEMENTARY MECHANICS. 321 

together, all would strike the ground at the same time. If the 
balls were melted together into one ^ 

ball, as b, they would still fall to- @ q/'q q 
gether, and strike the ground in the 
same time as before. ^^^* ^^* 

Since a number of horses cannot run a mile in less time 
than a single horse, so 100 pounds can fall no further in a 
given time than one pound can. 

894. Acceleration is the rate of change of velocity. 
If a force acts upon a body free to move, then, according to 
the first law of motion, it will move forever with the same 
velocity unless acted upon by another force. 

Suppose that, at the end of one second, the same force 
were to act again, the velocity at the end of the second 
second would be twice as great as at the end of the first 
second. If the same force were to act again, the velocity 
at the end of the third second would be three times that at 
the end of the first second. So, if a constant force acts 
upon a body free to move, the velocity of the body at the 
end of any time will be the velocity at the end of one second, 
multiplied by the number of seconds. 

895. This constant force is called a constant acceler- 
ating force, or constant retarding force, acccrdmg as 
the velocity is constantly increased or decreased. 

If a body is dropped from a high tower, the velocity with 
which it approaches the ground will be constantly increased 
or accelerated ; for the attraction of the earth, or force of 
gravity, is constant, and acts upon the body as a constant 
accelerating force. It has been found by careful experi- 
ments that this force of gravity, or constant accelerating 
force, on a freely falling body, is equivalent to giving the 
body a velocity of 32.16 feet in one second; it is always de- 
noted by g. As was mentioned before, g varies at difi'erent 
points of the earth, being 32.0902 at the equator, and 
32.2549 at the poles. Its value for this latitude (about 41^ 
25' north) is very nearly 32. IG, and this value should always 
be used in solving problems. It has also been found by 



3'^2 



ELEMENTARY MECHANICS. 



00 



00 








«0 




'tJ 




1^ 








%^ 


00 


V 


*« 


09 


s 


c^ 


o 




u 
<» 


%% 



«o 
«0 



Velocity at ^^ /'r\'« 
the end o/ ^ ( /' S 
one second §— * — ""^^ 
ia 32.16 feet ^ 



per second.'^ 

09 



Vbloeityai the 
end of two sec- 
onds is 64.32 
feet per seconds 



<1 



^ 









1 

^2 

*'3^ 



icHt 



#1' 

« 

I 

»4 



Velocity at the end of three a 
seconds is 96.48 feet per second. 



i 

3 



m 



-1 

-2 
'3 

-4 
-5 

-6 

-7 
•-8 



-^{9^-r-'S 



I 

I 



VUoeity at the end of four seconds 
is i28.64 feet per second. 



s 

.5 



•1 

-2 

■3 
4 
■8 

-6 
7 

-■8 

-9 

-10 

-11 
■12 
13 
14 
15 
4 



m 



Fig. 122. 



ELEMENTARY MECHANICS. 323 

experiment that a freely falling body starting from rest will 
have fallen 16.08 feet at the end of the first second; 64.32 
feet at the end of the second second; 144.72 feet at the end 
of the third second; 257.28 feet at the end of the fourth 
second, etc., all of which are shown in the diagram, Fig. 122. 
o- 64.32 , ^, 144.72 ^ ^, 257.28 ,^ ,„ 

^^^^^ 16X8 = ^ = ^ 'T6:0^=' = ^' -16:08-=^^ = ^^^ 

and 2^, 3^, 4^ are the squares of the number of seconds dur- 
ing which the body falls, it is easy to see that the space 
through which a body free to move will fall in a given time is 
equal to 16.08 multiplied by the square of the time in seconds. 

Of) -1 f* 

Since 16.08 = • — '^ — = |^, the space = |-^ x square of time 
in seconds. 

896. Formulas for Falling Bodies : — 

Let ^ = force of gravity = constant accelerating force due 
to the attraction of the earth ; 
/ = number of seconds the body falls; 
^ = velocity at the end of the time /; 
A = distance that a body falls during the time /. 
v = ^t, (13.) 

Tka^ is J tJie velocity acquired by a freely falling body at the 
end of t seconds equals 32. 16, multiplied by the time in seconds. 

Example, — What is the velocity of a body after it has fallen four 
seconds, assuming that the air offered no resistance ? 

Solution. — Using formula 13, 

z/ = ^/ = 32. 16 X 4 = 128. 64 feet per second. Ans. 

* = ~ (14.) 

That is, the number of seconds during which a body must 
have fallen to acquire a given velocity equals the given veloc- 
ity in feet per second, divided by 32. 16. 

Example. — A falling body has a velocity of 192.96 feet per second; 
how long had it been falling at that instant ? 
Solution. — Using formula 14, 

^ V 192.96 OAK 
i = — = ^^ ^„ = 6 seconds. Ans^ 
^ &>6. Id 



324 ELEMENTARY MECHANICS. 



v" 



h=^, (15.) 

That ts^ the height from which a body must fall to acquire 
a given velocity equals the square of the given velocity^ di- 
vided by 2 X S2,16. 

Example. — From what height must a stone be dropped to acquire a 
velocity of 24,000 feet per minute ? 

Solution. — 24,000 -r- 60 = 400 feet per second. Using formula 15, 
I, ^' 4002 160,000 cA^ri.af . A 

^ = ^ = 23^3216" = -6432- = ^'^^^'^^ ^^^^- ^^^• 

z/ = /^^ (16.) 

That is^ the velocity that a body will acquire in falling 
through a given height equals the square root of the product 
of twice S2. 16^ and the given height. 

ExAMPi.fi. — A body falls from a height of 400 feet; what will be its 
velocity at the end of its fall ? 

Solution. — Using formula 16, 

V = /s/^Tgh = '/2X 32.16x400 = 160.4 feet per second. Ans. 
h = \gt\ (17.) 

' That is, the distance a body will fall in a given time equals 
S2.16~2y multiplied by the square of the number of sec- 
ends. 

Example. — How far will a body fall in 10 seconds ? 
Solution. — Using formula 17, 

h = i^/2 = ^ X 32.16 X 102 = 1,608 feet. Ans. 



= ^/4^. (18.) 



g 
That is, the time it will take a body to fall through a given 

height equals the square root of twice tlie height divided by 

S2.16. 

Example, — How long will it take a body to fall 4,116.48 feet ? 
Solution. — Using formula 18, 

/ = 4/IZ5ll?3§: = 16 seconds. Ans. 
y 32.16 

897. A body thrown vertically upwards starts with a 
certain velocity called the initial velocity. In this case 
gravity acts as a constant retarding force. The formulas 
eiven above will also apply in this case. 



ELEMENTARY MECHANICS. 



325 



Example. — If a cannon ball is shot vertically upwards with an initial 
velocity of 2,000 feet per second, {a) how high will it go ? (J?) How 
'ong a time must elapse before it reaches the earth again ? 

Solution.— («) Using formula 15, 



y_V^_ 2,000^ 

27 ~ '2X32.16 



= 62,189 feet, nearly, = 11.778 miles. Ans. 



To find the time it takes to reach a height of 62,189 feet, use 
formula 14. 



/ = ^ 



2,000 



= 62.19 seconds. 



g 32.16 

Since it will take the same length of time to fall to the ground, the 
total time will be 62.19 X 2 = 124.38 seconds = 2 minutes 4.38 seconds. 
Ans. 



PROJECTILES. 

898. Any body thrown into the air is a projectile, and 

is acted upon by three forces — the original or initial force, 
the force of gravity, and the resistance of the air. We shall 
here consider only those projectiles which are thrown hori- 
zontally. 

899. The range is the horizontal distance between the 
starting point and the point where the body strikes the 
ground. In Fig. 123, sup- 
pose that A represents the 
starting point of the pro- 
jectile, and that it is shot 
horizontally outwards in 
the direction of the arrow 
with a velocity of 70 feet 
per second. Now, if the 
resistance of the air be 
neglected, the velocity in 
the horizontal direction will 
be uniform, and the pro- 
jectile will pass over equal 
spaces in equal times. Let 
A 1 represent 70 feet, or 
the space passed over in one second. At the end of five 
seconds, if gravity had not acted upon the projectile, it 




402 

Fig. 123. 



D. 0. III. 



r 



326 ELEMENTARY MECHANICS. 

would have been at o, but as gravity has acted, it falh 
16.08 feet the first second; at the end of the second second 
it has fallen 64.32 feet, etc. 

Let A b represent the fall in one second — that is, 16.08 
feet, drawn to the same scale as A i, which represents 70 
feet. Now, complete the parallelogram A 1 B b^ and B will 
be the point which the projectile has reached at the end of 
one second. \i A c represents 64.32 feet, and the parallelo- 
gram A 2 C c\^ completed, the projectile will be at C at the 
end of the second second. Proceeding in this manner, find 
the points D, E, and F, the positions of the projectile at 
the end of 3^ ^ and 5 seconds, respectively. Drawing the 
curve A B C D E F through the points thus found, it rep- 
resents the path of the projectile. This curve is called 
a parabola. 

The distance H F is the range^ and, as is easily seen, 
equals the time in seconds multiplied by the original velocity 
in feet per second. 

900. If the height A H and the initial velocity are 
given, and it is desired to find the range H F^ calculate the 
time that it will take to fall through a height equal to the 
given height^ and multiply the time thus found by the initial 
velocity. 

Example. — A cannon ball is fired in a horizontal direction with an 
initial velocity of 1,500 feet per second. If the mouth of the cannon is 
25 feet above the ground, what is its range ? 

Solution. — Applying formula 18, Art. 896, 

^ ,/2^ , /2 X 25 . ^.„ , . 

t ■=■ \/ — = V ..^ ^^ = 1.247 seconds, nearly. 

y g ^ oJi.lO 

Range ^vt = 1,500 X 1.247 = 1,870.5 feet. Ans. 

Example. — A projectile has an initial velocity of 90 feet per second. 
If it is desired to strike an object 15 feet away, how far below the hori- 
zontal line of direction must the object be located ? 

Solution. — The object must be located as far below as the distance 
that the body would fall, through the action of gravity, during the 
time it would take in passing over a distance of 15 feet at a velocity of 
90 feet per second. 

Hence, 15 -j- 90 = ^ of a second. Applying formula 17, Art. 896, 

/5 = ^^/2 = ^ X 32.16 X (i/ = .447 foot, nearly, = 5.36 inches. Ans. 



ELEMENTARY MECHANICS. 327 

EXAMPLES FOR PRACTICE. 

1. A body starts from a state of rest, and falls freely for nine sec- 
onds; how far will it fall ? Ans. 1,302.48 ft. 

2. What velocity must a body have in order to carry it upwards 500 
feet, vertically ? Ans. 179.33 ft. per sec. 

3. A baseball is thrown vertically upwards to a height of 200 feet ; 
how long a time must elapse before it strikes the ground ? 

Ans. 7.05 sec. 

4. What will be the velocity of a freely falling body at the end of 6 
seconds? Ans. 192.96 ft. per sec. 

5. A baseball is thrown horizontally 5 feet above the ground, with 
a velocity of 80 feet per second; what is its range ? Ans. 44.61 ft. 

6. A leaden bullet falls from a tower 100 feet high; with what 
velocity will it strike the ground ? Ans. 80.2 ft. per sec. 

7. A bullet is dropped from a high tower. If it takes 4^ seconds to 
reach the ground, how high is the tower ? Ans. 290.445 ft. 

8. A freely falling body has a velocity of 400 feet per second ; how 
long has it been falling ? Ans. 12.438 sec 

CENTRIFUGAL FORCE. 

901. If a body be fastened to a string and whirled so as 
to give it a circular motion, there will be a pull on the string, 
which will be greater or less according as the velocity in- 
creases or decreases. The cause of this pull on the string 
will now be explained. 

Suppose that the body is revolved horizontally, so that the 
action of gravity upon it will always be the same. According 
to the first law of motion, a body put in motion tends to 
move in a straight line unless acted upon 
by some other force, causing a change in .X 
the direction. When a body moves in a / '\ 

circle the force that causes it to move in \ j 

a circle instead of a straight line is ex- \ ./ 

actly equal to the tension of the string. '^^-. ^''^ 

If the string were cut, the pulling force ig. i . 

that drew it away from the straight line would be removed 
and the body would then *' fly off at a tangent" — that is, it 
would move in a straight line tangent to the circle, as 
shown in Fig. 124. 

902. Since, according to the third law of motion, every 
action has an equal and opposite reaction, we call that force 



328 ELEMENTARY MECHANICS. 

which acts as an equal and opposite force to the pull of the 
string the centrifugal force, and it acts away from the 
center of motion. 

903. The other force or tension of the string is called 
the centripetal force, and it acts towards the center of 
motion. It is evident that these two forces acting in oppo- 
site directions tend to pull the string apart, and, if the 
velocity be increased sufficiently, the string will break. It 
is also evident that no body can revolve without generating 
centrifugal force. The value of the centrifugal force of any 
revolving body, expressed in pounds, is 

F= .00034 W R N\ (19.) 
in which F = centrifugal force ; 

JV= total weight of body in pounds; 

R = radius, usually taken as the distance 
between the center of motion and the 
center of gravity of the revolving body, 
in feet; 

iV= number of revolutions per minute. 

904. In calculating the centrifugal force tending to 
burst a fly-wheel, it is the usual practice to consider one- 
half the rim of the wheel only, and not to take the arms and 
hub of the wheel into account. In this case, R is taken as 
the distance between the inside edge of the rim and the center 
of the shaft and the whole is divided by 3.1416. 

Note. — The general formula for centrifugal force is F-= — q-, where 

K 

m — the mass of the revolving body, v = velocity of center of gravity 

of body in feet per second, and i^ = radius, as above. Formula 19, 

IV 60 V 

Art. 903, is easily derived from this. Thus: m = — ; N= jr — ^, or 



^ = -60- ' ^^"^^' ^^ -R ^ =i^V~60-J = -^^^^^ ^^ ^ • 

Example. — What would be the centrifugal force tending to burst a 
cast-iron fly-wheel whose outside diameter was 10 feet, width of face, 
20 inches, and thickness of rim 6 inches, turning at the rate of 80 revo- 
lutions per minute ? 

Solution. — First calculate the weight of one-half the rim. The 
diameter of the rim = 10 X 12 = 120 inches; the diameter of the cir- 
cle midway between the inside and outside diameters of the rim = 



ELEMENTARY MECHANICS. 



329 



120 — 6 = 114 inches. The number of cubic inches in the rim = 114 X 
3.1416 X 20 X 6 = 42,977 cubic inches. 42,977 X .261 X i = 5,608.5 
pounds = weight = ^. 7? = i^ — i = 4^ f eet. N=SO. 



Hence, J^= .00034 W R N'' ■ 
3.1416 = 17,481+ pounds. Ans. 



3.1416 = .00034 X 5,608.5 X 4i X 80^ ^ 



STATICS. 
905. Statics may be defined as that branch of Me- 
chanics which treats of bodies at rest or of bodies moving 
with a uniform velocity, when these bodies are acted upon 
by forces. A body is in static equilibrium when the 
resultant of all of the forces acting upon the body is zero. 




Fig. 125. 



MOMENTS OF FORCES, 

906. If from any point (9, Fig. 
125, a perpendicular be drawn to the -p 
line of action of a force, the product 
of the magnitude of the force and 
the length of the perpendicular is 
called the moment of ttie force 
about the point O. 

Thus, in the figure, the moment of 
the force F' about the point O is F' 
X O B; oi the force F" about the point O is F" X O Ay 
and of F'" is F'" X O C. 

907. The use of the moment will be explained further 
on, when the necessity arises for using it. The point O is 
called the center of moments. 

908. When two equal forces act in parallel lines, but in 
opposite directions, they constitute what is called a couple. 

909. In Fig. 12G, the equal and parallel forces F^ and 

|1^ F\ acting in opposite directions 

(one up and the other down), form 

a couple. It is easy to see that if 

they were joined by a connectioi:i, 

I 2.^ A B^ that they would tend to 

I ^ turn A B about the point C, midway 

Fig. 126. between F' and F", The moment 



330 



ELEMENTARY MECHANICS. 



of a couple about any point is always the same, and is equal 
to the product of one of the equal forces into the perpendicular 
distance between the two forces. 

Thus, the moment of the couple in the figure equals F' or 
F" multiplied by A B. An example of a couple would be a 
wrench applied to a nut. Here, two opposite and parallel 
sides of the wrench act in parallel, but opposite, directions, 
against two parallel sides of the nut. 



CENTER OF GRAVITY. 

91 0. The center of gravity of a body is that point at 
which the body may be balanced^ or it is the point at which 
the whole weight of a body may be considered as concentrated. 

In a moving body, the line described by its center of 
gravity is always taken as the path of the body. In finding 
the distance that a body has moved, the distance that the 
center of gravity has moved is taken. 

911, The definition of the center of gravity of a body 
may be applied to a system of bodies, if they are considered 
as being connected at their centers of gravity. 

If w and PF, Fig. 127, be two bodies of known weights, 
their center of gravity will be at C. The point C may be 

, readily determined, as 
follows : Take C as the 
center of moments ; 
then, since the weights 
are to balance each 
other, the moment of 
W about C must equal 
the moment of w about 
C\ or, in other words, 
IV X C W= w X Cm. 




Fig. 137. 




If the distance between the centers of gravity of W and w 
IS known, it is very easy to find C w and C W, For 



ELEMENTARY MECHANICS. 



331 



Let /=the distance wlV between the centers of the 
bodies ; 

/j = the short arm CW ; 
w — weight of small body ; 
W = weight of large body. 
Then, since wC = I — l^, we have, taking the moments 
about the point C^ 

Wl^ = w{l — /j) z= wl — wl^ ; whence, 
Wl^ '\-wl^ = ( W~\- w)l^ = w/, or 

A = -J|q^- (20.) 

Example. — In Fig. 127, w = \0 pounds, ^=30 pounds, and the 
distance between their centers of gravity is 36 inches; where is the 
center of gravity of both bodies situated ? 

Solution. — Applying formula 20, 

wl 10 X 36 



CW=h = 



= 9 inches ; 



lV+7£/- 30 + 10 

Aence, the center of gravity is 9 inches from the center of the larger 
body. Ans. 

The general method for finding the short arm CW is, 
then, as follows : Multiply the iveight of the smaller body by 
the distance between tJie centers of the two bodies^ and divide 
this product by the sum of the weights of the two bodies. 

912. It is now very easy to extend this principle, to 
the finding of the center 



of gravity of any num- 
ber of bodies when their 
weights and the dis- 
tances apart of their 
centers of gravity are 
known, by applying 
the principle of finding 
the resultant of several 
forces; that is, by find- 
ing the center of grav- 
ity of two of the bodies, 
as W^ and W^^ in Fig. 
128. at C,. 




Pig. 128. 



332 



ELEMENTARY MECHANICS. 



Assume that the weight of both bodies is concentrated at 
C^y and find the center of gravity of this combined weight 
at C^ and of W^ to be at C^ ; then, find that the center of 
gravity of the combined weights of W^^ W^, and W^ (concen- 
trated at CJ and IV^ to be at Cy and C will be the center of 
gravity of the four bodies. 

913. To find the center of 

gravity of any parallelogram : Draw 

the two diagonals^ Fig. 129, and their 

point of intersection C will be the 

FIG. 129. ^^/^/^r of gravity. 

914. To find the center of gravity of a triangle, as 
A B Cy Fig. 130 : From any ver- 
tex, as A , draw a line to the mid- 
dle point D of the opposite side B C. 
From one of the other vert exes, as C, 
draw a line to F, the middle point 
of the opposite side A B ; the point 
of intersection, O, of these two lines, is the center of gravity. 






Fig. 131 



It is also true that the distance D O = ^ D A, and that 



ELEMENTARY MECHANICS. 



333 



F O = ^ F C^ and the center of gravity could have been 
found by drawing from any vertex a line to the middle 
point of the opposite side, and measuring back from that 
side ^ of the length of the line. 

The center of gravity of any regular plane figure is the 
same as the geometrical center. 

915. To find the center of gravity of any irregular 
plane figure, but of uniform thickness throughout, divide 
one of the parallel surfaces into triangles, parallelograms, 
circles, ellipses, etc., according to the shape of the figure ; 
find the area and center of gravity of each part separately, 
and combine the centers of gravity thus found, as in the 
case of more than two bodies whose weights were known, 
except that the area of each part is used instead of their 
weights. See Fig. 131. 

916. Center of Gravity of a Solid. — In a body free 
to move, the center of gravity will lie in a vertical plumb 




Fig. 132. 
line drawn through the point of support. Therefore, to 
find the position of the center of gravity of an irregular 
solid, as the crank, Fig. 132, suspend it at some point, as B^ 
so that it will move freely. Drop a plumb line from the 
point of suspension, and mark its direction. Suspend the 
body at another point, as A^ and repeat the process. The 



334 ELEMENTARY MECHANICS. 

intersection of the two lines will be directly over the center 
of gravity. 

Since the center of gravity depends wholly upon the 
shape and weight of a body, it may be without the body, as 
in the case of a circular ring, whose center of gravity is at 
the center of the circumference of the ring. 



EQUILIBRIUM. 

917. When a body is at rest, all of the forces which act 
upon it are said to balance one another, or to be in equilib- 
rium. The most important of the forces is gravity, 
which acts upon every molecule of the body. 

918. There are three states of equilibrium : Stable, 
unstable, and neutral. 

919. A body is in stable equilibrium when, if 
slightly displaced from its position of rest, it tends to return 
to that position. 

For example, a cube, a cone resting on its base, a pendu- 
lum, etc. 

If a body is in stable equilibrium, its center of gravity 
is raised when it is displaced. 

920. A body is in unstable equilibrium when, if 
slightly displaced from its position of rest, it tends to fall 
farther fr 0171 that position. 

For example, a cone standing upon its point, an egg 
balanced upon its end, etc. 

Any movement, however slight, lowers the center of 
gravity when the body is in unstable equilibrium. 

921. A body is in neutral equilibrium when it has 
no tendency to move either way., in the direction of its mo- 
tion, after being slightly displaced. 

For example, a sphere of uniform density ; a cone resting 
on its side. 

922. A vertical line drawn through the center of gravity 
of a body is called the line of direction. So long as the line 
of direction falls within the base, the body will stand. When 
the line of direction falls without the base, the body will fall. 



ELEMENTARY MECHANICS. 



335 







Let A C Bf Fig. 133, be a cylinder whose base is oblique 
JB js to the center line 

B O n ; and let O 
be the center of 
gravity of this cyl- 
inder. 

So long as the 

pe rpendicular 

c A _^ C through O falls be- 

FiG. 133. between A and (T, 

the cylinder will stand, but the instant that it falls without 

the base, the cylinder will fall. 

T/ie center of gravity of a body always tends to seek its 
lowest point. 

EXAMPLES FOR PRACTICE. 

1. There are three weights in a straight line. The first weighs 
40 lb. ; the second, 16 lb., and the third, 50 lb. Distance between the first 
and second is 6 ft., and between the second and third, 10 ft. Where is 
the center of gravity ? Ans. 8 ft. 5.434 in. from the 40-lb. weight. 

2. Find the perpendicular distance between the center of gravity 
and the longer side of a triangle whose sides are 7 ft., 10 ft., and 15 ft. 
long. Solve graphically. Ans. 1.31 ft. 

3. A rectangle, 2 ft. long and 1 ft. wide, has equal weights of 50 lb. 
each, suspended from two of its diagonally opposite corners. A weight 
of 60 lb. and another of 80 lb. are suspended from the other two corners. 
Supposing the rectangle to be without weight, where is the center of 
gravity ? a ( On the diagonal joining the 60-lb. and 

' ( 80-lb. weights, 1.118 in. from the center. 

4. Find the center of gravity of a quadrilateral whose sides are 14, 
15, 16, and 18 in. long, the angle between the 18 and 16 in. sides being 
45°. Give the perpendicular distance from the 18 in. side, Ans, 5,46 in. 



SIMPLE MACHINES. 



THE LEVER. 

923. A lever is a bar capable of being turned about a 
pin, pivot or point, as in Figs. 134, 135 and 136. 

924. The object W to be lifted is called the weight; 
the force used P is called the po^ver ; and the point or 
pivot F is called the fulcrum. 



336 



ELEMENTARY MECHANICS. 



925. That part of the lever between the weight and 
the fulcrum, or F b, is called the iveight arm, and the 
part between the power and the fulcrum, or F c^ is called 
the poiiver arm. 

926. Take the fulcrum, or point i% as the center of 
moments; then, in order that the lever shall be in equilib- 
rium, the moment of P about i% or Px F c^ must equal the 
moment of W about i% or Wx F b. That is, Px Fc = 
Wx F b, or, in other words, t/ie power multiplied by the 

power arm equals the wcigJit multiplied by the weight ar7n. 

927. If /^be taken as the center of a circle, and arcs be 
described through b and <:, it will be seen that, if the weight 
arm is moved through a certain angle, the power arm will 
move through the same angle; also, that the vertical dis- 
tance that W moves will be the sine of this angle, in a 
circle whose radius is the weight arm, and that the vertical 
distance that P moves will be the sine of the same angle in 
a circle whose radius is the power arm. From this it is seen 
that the power arm is proportional to the distance through 



fP 



IJP 



t£_ 



Pig. 134. 



Fig. 135. 



Pig. 136. 



which the power moves, and the weight arm is proportional 
to the distance through which the weight moves. 

Hence, instead of writing PxF c^= WxF b^ we might 
have written it P X distance through which P moves =z JVx 
distance through which ^F moves. This is the general law 
of all machines, and can be applied to any mechanism, from 
the simple lever up to the most complicated arrangement. 
Stated in the form of a rule, it is as follows : 

Rule VI. — The power multiplied by the distance through 
which it moves equals the weight multiplied by the distance 
through which it m,oves. 



ELEMENTARY MECHANICS. 



337 



Example. — If the weight arm of a lever is 6 inches long and the 
power arm is 4 feet long, how great a weight can be raised by a force 
of 20 pounds at the end of the power arm ? 

Solution. — 4 feet = 48 inches. Hence, 20x48= JVx^, or py = 
160 pounds. Ans. 

Example. — (a) What is the ratio between the power and the weight 
in the last example ? (d) In the last example, if F moves 24 inches, 
how far does IV move ? (r) What is the ratio between the two distances? 

Solution. — (a) 20 : 160 = 1:8; that is, the weight moved is 8 times 
the power. Ans. 

(d) 20 X 34 = 160 X -r. .r = -^ = 3 inches, the distance that IV 



moves. Ans. 



160 



(^) 3 : 24 = 1 : 8, or the ratio is 1 : 8. Ans. 

928. The law which governs the straight lever also 
governs the bent lever; but care must be taken to deter- 
mine the true lengths of the lever arms, which are in every 
case the perpendicular distances from the fulcrum to the line 
of direction of the weight or power. 

Thus, in Figs. 137, 138, 139, and 140, F c \n each case 
represents the power arm, and F b the weight arm. 
c F h 




Fig. 139. Fig. 140. 

929. The Compound Lever. — A compound lever is 

a series of single levers arranged in such a manner that 



338 ELEMENTARY MECHANICS. 

when a force is applied to the first it is communicated to 
the second, and from this to the third, and so on. 

Fig. 141 shows a compound lever. It will be seen that 
when a power is applied to the first lever at P it will be 
communicated to the second lever at P, from this to the 
third lever at P, and thus raise the weight W. 

The weight which the power of the first lever could raise 
acts as the power of the second, and the weight which this 
could raise by means of the second lever acts as the power 
of the third lever, and so on, no matter how many single 
levers make up the compound lever. 

In this case, as in every other, the power multiplied by 
the distance through which it moves equals the weight 
multiplied by -the distance through which it moves. 

Hence, if we move the P end of the lever, say, 4 inches, 
and the W end m.oves \ of an inch, we know that the ratio 

F 



I T 



o-t 



—24^ ^ 6^ — IS—-^ 6 ^ -30^^ f—- jy ^^ 

w\ 



^ TV P A 

F F 

Fig. 141. 

between Pand Wis the same as the ratio between ^ and 4; 
that is, 1 to 20, and, hence, that 10 pounds at P would 
balance 200 pounds at JV^ without measuring the lengths of 
the different lever arms. If the lengths of the lever arms 
are known, the ratio between P and W may be readily 
obtained from the following rule: 

Rule VII. — The continued product of the power and each 
power arm equals the continued product of the weight and 
each weight arm. 

Example. — If, in Fig. 141, the power arms P F= 24 inches, 18 inches, 
and 30 inches, and weight arms W F=- 6 inches, 6 inches, and 18 inches, 
{a) how great a force must be applied at the free end P to raise 1,000 
pounds at Wl (<^) What is the ratio between /* and Wl 

Solution.— P x 24 x 18 X 30 = 1,000 x 6 x 6 x 18, 

-, 648,000 r,c^ A K 

or P= .^ -,_„ = 50 pounds. Ans. 
50 : 1,000 =^ 1 : 20, or P : IV= 1 : 20. An& 



1 



ELEMENTARY MECHANICS. 



339 



THE -WHEEL AND AXLE. 

930. The ^vlieel and axle consists of two cylinders of 
different diameters^ rigidly connected^ so that they turn 
together about a common axis, as in Fig. 142. Then, as 




I 



Fig. 142. 

before, /*x distance through which it moves = W^x dis- 
tance through which it moves; and, since these distances 
are proportional to the radii of the power cylinder and 
weight cylinder, Py^ F c— W Y, F b. 

It is not necessary that an entire wheel be used ; an arm, 
projection, radius, or anything ^= 
which the power causes to revolve 
in a circle, may be considered as 
the wheel. Consequently, if it is 
desired to hoist a weight with a 
windlass. Fig. 143, the force is 
applied to the handle of the crank, 
and the distance between the cen- f^g. 143. 

ter line of the crank-handle and the axis of the drum corre- 
sponds to the radius of the wheel. 

Example. — If the distance between the center line of the handle and 

the axis of the drum, in Fig. 143, is 18 inches, and the diameter of the 

drum is 6 inches, what force will be required at P to raise a load of 300 

pounds ? 

/» 

Solution.— P X 18 = 300 x o-» ov P = 50. Ans. 




340 



ELEMENTARY MECHANICS. 



931 • Wlieel'work. — A combination of wheels and axles, 
as in Fig. 144, is called a train. The wheel in a train to 
which motion is imparted from a wheel on another shaft, 
by such means as a belt or gearing, is called the driven 
wheel or follower ; the wheel which imparts the motion 
is called the driver. 




Fig. iU. 



932. It will be seen that the wheel and axle bears the 
same relation to the train that the simple lever does to the 
compound lever; that is, tAe continued product of the power 
and the radii of the driven wheels equals the continued product 
of the weighty the radius of the drum that moves the weighty 
and the radii of the drivers. 

Example. — If the radius of the wheels is 20 inches, of C, 15 inches, 
and of E, 24 inches : if the radius of the drum 7^ is 4 inches, of the 



ELEMENTARY MECHANICS. 341 

pinion Z>, 5 inches, and of the pinion B, 4 inches, how great a weight 
will a force of 1 pound at P raise ? 

7 200 
Solution.— 1 x 20 X 15 X 24 = J^X 4 x 5 x 4, or W = -^ = 90 

pounds. Ans. 

933. Hence, also, if PFwere raised one inch, P would 
move 90 inches, or P would have to move 90 inches to raise 
J/Fone inch. It is now clear that another great law has 
made itself manifest, and that is that, whenever there is a 
gain in power without a corresponding increase in tJie initial 
force^ there is a loss in speed. 

In the last example, if /'were to move the entire 90 inches 
in one second, PF would move only 1 inch in one second. 
The same principle may be applied to any machine. 



THE PULLEY. 

934. A pulley is a wheel turning on an axle, over which 
a cord, chain, or band is passed in order to transmit the 
force through the cord, chain, or band. 

935. The frame which supports the axle of the pulley 
is called the block. 



936. A fixed pulley is one whose block 
is not movable, as in Fig. 145. In this case, 
if the weight W be lifted by pulling down P, 
the other end of the cord W will evidently 
move the same distance upwards that /'moves 
downwards ; hence, P must equal W, 




Fig. 145. 



937. A movable pulley is one whose block is movable, 
as in Fig. 148. One end of the cord is fastened to the beam, 
and the weight is suspended from the pulley, the other end 
of the cord being drawn up by the application of a force P. 
A little consideration will show that if P moves through a 
certain distance, say 1 foot, JFwill move through halfXhdX 

D. 0. JlL—.i, 



'? 



342 



ELEMENTARY MECHANICS. 



distance, or 6 inches; hence, a pull of 1 pound at Pwill lift 
2 pounds at W. 





Fig. 146. Fig. 147. 

The same would also be true if the free end of the cord 
were passed over 3. fixed pulley^ as in Fig. 147, in which case 
the fixed pulley merely changes the direction in which P 
acts, so that a weight of 1 pound hung on the free end of the 
cord will balance 2 pounds hung from the movable pulley. 

938. A combination of pulleys, as shown in Fig. 
148, is sometimes used. In this case there 
are three movable and three fixed pulleys, 
and the amount of movement of W^ owing to 
a certain movement of P, is readily found. 

It will be noticed that there are six parts 
of the rope, not counting the free end ; hence, 
if the movable block be lifted 1 foot, P re- 
maining in the same position, there would 
be 1 foot of slack in each of the six parts of 
the rope, or six feet in all. Therefore, P 
would have to move 6 feet in order to take 
up this slack, or Amoves six times as far as W, 
Hence, 1 pound at Pwill support 6 pounds 
at W^ since Xh^ power multiplied by the dis- 
tance through which it moves equals the weight 
multiplied by the distance through which it 
moves. It will also be noticed that there are 
Pig. 148. three movable pulleys, and that 3x2 = 6. 




ELEMENTARY MECHANICS. 



343 



Rule VIII. — In any combination of pulleys where one con- 
tinuous rope is used a load on the free end will balance a 
weight on the movable block as many times as great as the 
load on the free end as there are parts of the rope supporting 
the load — not counting the free end. 

The above law is good, whether the pulleys are side by- 
side, as in the ordinary block and tackle^ or whether they are 
arranged as in the figure. 

Example. — In a block and tackle having five movable pulleys, how 
great a force must be applied to the free end of the rope to raise 1,250 
pounds ? 

Solution. — Since there are 5 movable pulleys, there are 10 parts of 
the rope supporting them, and 1 pound on the free end will balance 10 
pounds on the movable block ; therefore, the ratio of /* to f^ is 1 : 10, 



1 250 
and P = ' =125 pounds. 



Ans. 



iJL. 






I 



939. In Fig. 149 is shown an arrangement called a dif- 
ferential pulley. It will be seen that if a force be applied 
at P^ so as to pull the point P down to D^ the ^ 

rope or chain will wind up on the large pulley 
A^ and unwind from the smaller pulley B^ 
and since (T is a movable pulley, the weight W 
will move an amount equal to one-half the 'vV^^.^ 
difference between the amount of winding I jj*^ — ^ 
on A and unwinding on B. 

Let the radius of A be represented by R^ 
and of B hy r\ then, when P is pulled down 
to D^ a point E on the pulley A will move 
through a certain angle, EOF, the length of 
the arc E 7^ being equal to the distance P D. 
Any point on the pulley B, which is fastened to 
A, will turn through the same angle; and the 
difference between the arc^Z, through which 
this point turns, and the arc E F will be pro- 
portional to the difference of the radii R and r. 

When P moves down to D, the point // on 
theother side will move up' through the same distance to//"'. 
Thepoint/ will move up to /', a distance equal to the length 




Fig. 149. 



344 



ELEMENTARY MECHANICS. 



of the arc K L, and m will fall through the same distance 
to 7n\ thus causing the weight W to be raised one half the 
difference betzveen m m' and P D. The ratio between the 
distances through which /Kand Pmove will be 



arc E F — arc K L 



Hence, W y^ 



2 
R-r 

2 



E F^ or 



R 



PxR, or IV = 



2 

2 PR 
R-r 



R, 



(21.) 



Example. — If i? = 7 inches, and r = 6^ inches, how much weight can 
be raised at IV with a force of 50 pounds at jP ? 



Solution.- 



JV = 



2FR 2x50x7 



J^ - r 



7-6i 



= 1,400 pounds. Ans. 



THE INCLINED PLANE. 

940. An inclined plane is ^ slope or a flat surface, 
making an angle with a horizontal line. 

Three cases may arise in practice with the inclined plane: 

I. When the power acts parallel to the plane, as in Fig. 
150. 

II. When the power acts parallel to the base, as in Fig. 
151. 

III. When the power acts at an angle to the plane, or 
to the base, as in Fig. 152. 

Case I.' — In Fig. 150, the relation existing between the 
power and the weight is easily found. The weight ascends 
a distance equal to c b^ or the height of the inclined plane, 

while the power descends 
through a distance equal 
to a b^ or the length 
of the inclined plane. 
Therefore, the power 
multiplied by the length 
of the inclined plane 
Fig. 150. equals the weight multi- 

plied by the height of the incliiied plane. Hence, if the 
length ab=.^:<^ feet, and the height c b — "l^ feet, /^ X 20 =r 
P X 40, or 1 pound at P will balance 2 pounds at W~ 




^ lbs. 



ELEMENTARY MECHAStICS. 



345 




1 lb. 



Fig. 151. 



Case II. — In Fig. 151, the power is supposed to act par- 
allel to the base for any position of W, the pulley being 

shifted up and down; 
therefore, while W is 
moving from the level 
a c to b, or through the 
height c b oi the in° 
clined plane, P will 
move a distance down- 
wards, relative to the 
axis of the pulley, equal 
to the length of the 
base a c. Hence, when the power acts parallel to the base, 
W X height of the inclined pla7ie = P x length of base. 

If the length of the base is 40 feet, and the height of the 
inclined plane is 20 feet, 1^ X 20 = P X 40, and 1 pound at 
P will balance 2 pounds at W, 

Case III. — For Fig. 
152 no rule can be given. 
The ratio of the power 
to the weight must be 
determined for every 
position of l^by means 
of the triangle or paral- 
lelogram of forces. 

This case will be ex- 
plained by means of an 

example, as it affords a fig. 152. 

splendid illustration of the principle of resolution of forces. 

Example. — In Fig. 153, a body W is shown resting on an inclined 
plane A B, whose dimensions are marked on the cut ; the weight P acts to 
pull the body up the plane by means of the rope r and pulley p. It 
is required to find what the weight of P must be in order to start IV up 
the plane. Suppose M^ weighs 120 pounds, and that friction is neglect- 
ed. It is also required to find the perpendicular pressure which IV 
exerts against the plane. 

Solution. — Through the point a, the center of gravity of W, draw 
a b vertical, and make it of such a length as to represent 120 pounds 
to a convenient scale, say 60 pounds = 1 inch. Drawing ac and cb^ 




346 



ELEMENTARY MECHANICS. 



respectively, parallel and perpendicular to the plane, a c represents 

the magnitude of the force which must be exerted parallel to ^ ^ in 

order to put the body in equilibrium — i. e. , to balance the force which 

gravity exerts in pulling the body down the plane. If the rope r were 

parallel to A B, ac would represent the 

weight of P\ but, since r makes an angle 

with the plane, Pwill not be equal to X / P 

a c. To find what the weight of P must 

be, draw a d parallel to a c, but indicate 

it as acting in the opposite 

direction, or from a to d in- ^^ 

stead of from a to c. Now 




treat ad as though it were a compo- 
nent of the force acting in the rope — 
i. e., draw de perpendicular to a d, in- 
stead of perpendicular to a e. The 
reason for this is that ii d e were 
drawn perpendicular to ae it could 
be resolved into components, one of 
which would be parallel to ad, 3. re- 
sult which we wish to avoid ; in other Fig. 153. 
words, we want de perpendicular to the plane. The line ae, measured 
to the same scale as a b, will give the value of P. Measuring it, its 
length is .89" ; hence, P = .89 X 60 = 53.4 pounds. Ans. 

To determine the perpendicular pressure against the plane, it will 
be noticed that a b equals the pressure due to gravity. Since c b and 
de are both perpendicular to A B, they are parallel, and since de acts 
in the opposite direction to cb, the actual pressure against the plane is 
given by the difference between cb and de. Making ^y equal to d e, 
f b represents the perpendicular pressure against the plane when the 
force P {= ae) acts as shown. The length oi f b is 1.39" ; hence, the 
perpendicular pressure is 1.39 x 60 = 83.4 pounds. Ans. 

Since ca and ad are parallel and equal, and cf and de are also 
parallel and equal, it follows that af and a e must also be parallel and 
equal. Consequently, the force P might have been found by drawing 
a f parallel to the direction in which the pull on the rope acts, and 
bf perpendicular to the plane A B. Thus, suppose that the weight 
occupies the position shown by the dotted lines. Then, drawing ag 



ELEMENTARY MECHANICS. 



347 



parallel to a' e\ ag represents the weight of P, and g b represents the 
perpendicular pressure of the body W against the plane. Measuring 
ag, its length is .79" ; hence, /» = ,79 X 60 = 47.4 pounds. Measuring 
gby its length is 1.65" ; hence, the perpendicular pressure = 1.65 X 60 = 
99 pounds. 

941. The Wedge.— 
The ^vedge is a movable 
inclined plane, and is used 
for moving a great weight a 
short distance. A common 
method of moving a heavy 
body is shown in Fig. 154. 

Simultaneous blows of equal force are struck on the heads 
of the wedges, thus raising the weight VV. The laws for 
wedges are the same as for Case II of the inclined plane. 




Fig. 154. 





Fig. 1.55. 



THE SCREW. 

942. A screw 

is a cylinder with a 
spiral groove winding 
around its circumfer- 
ence. This spiral ia 
called the thread of 
the screw. The dis- 
tance that the thread 
is drawn back or ad- 
vanced in one turn of 
the screw is called the 
pitch of the screw. 
The screw in Fig. 155 is turned in a 
nut a by means of a force applied at 
the end of the handle P. For one com- 
plete revolution of the handle, the 
screw will be advanced lengthwise an 
amount equal to the pitch. If the nut 
be fixed, and a weight be placed upon 
the end of the screw, as shown, it will 
be raised vertically a distance equal 
to the pitch by one revolution of the 



348 ELEMENTARY MECHANICS. 

screw. During this revolution the force at P will move 
through a distance equal to the circumference, whose radius 
is PF. Hence, H^X pitch of thread = P X circumference 
of P. 

943. Single-threaded screws of less than 1 inch pitch 
are generally classified by the number of threads they have 
in 1 inch of their length. In such cases, one inch divided by 
the number of threads equals the pitch ; thus, the pitch of a 
screw that has 8 threads per inch is ^' ; one of 32 threads 
per inch is ^V, etc. 

Example. — It is desired to raise a weight by means of a screw having 
5 threads per inch. The force applied is forty pounds at a distance of 14 
inches from the center of the screw ; how great a weight can be raised ? 

Solution. — Diameter of the circumference passed through by the 
force = 14 X 3 = 28 inches. Therefore, ^ X i = 40 X 38 X 3. 1416, or 
W= 17,593 pounds. Ans. 

VELOCITY RATIO. 

944- The ratio of the distance that the power moves to 
the distance which the weight moves on account of the 
movement of the power is called the velocity ratio. 

Thus, if the power is moving 12 inches while the weight is 
moving 1 inch, the velocity ratio is 12 to 1, or 12; that is, 
P moves 12 times as fast as W. 

945. If the velocity ratio is known, the weight which 

any machine can raise equals the power 'inultiplied by the 

velocity ratio. If the velocity ratio is 8.7 to 1, or 8.7, H^ = 

8.7 X /", since I^X 1 = /^X 8.7. 

Note. — In all of the preceding cases, including the last, friction has 
been neglected. 

FRICTION. 

946. Friction is the resistance that a body meets from 
the surface on which it moves. 

947. The ratio between the resistance to the motion 
of a body due to friction and the perpendicular pressure be- 
tween the surfaces is called the coefficient of friction. 



ELEMENTARY MECHANICS. 



349 



If a weight W, as in Fig. 156, rests upon a horizontal 

plane, and has a cord fastened to it passing over a pulley a^ 

from which a weight Pis suspended, then, if Pis just suffi- 

P 
cient to start W, the ratio of P to PF, or -yp:, is the coefficient 

of friction between W and the surface it slides upon. 

The weight W is the perpendicular pressure, and P is the 
force necessary to overcome the resistance to the motion of 
W^due to friction. 





' 


1 






100 lbs. 
W 






(%^ 


^-^ 



10 lis. 



Fig. 156. 




If IV= 100 pounds and P= 10 pounds, the coefficient of 

P 10 

friction for this particular case would be -777 = ■ = .1. 

W 100 

948. Laws of Friction : — 

I. Friction is directly proportional to tJie perpendicu- 
lar pressure between the two surfaces in contact. 

II. Friction is independent of the extent of the surfaces 
in contact when the total peri)endicular pressure remains the 
same. 

III. Friction increases with the roughness of the sur- 
faces. 

IV. Friction is greater between surfaces of the same 
material than between those of different materials. 

V. Friction is greatest at the beginning of motion. 

VI. Friction is greater between soft bodies than be- 
tween hard ones. 

VII. Rolling friction is less than sliding friction. 

VIII. Friction is diminished by polishing or lubricating 
the surfaces. 



350 



ELEMENTARY MECHANICS. 



Law I shows why the friction is so much greater on jour- 
nals after they begin to heat than before. The heat causes 
the journal to expand, thus increasing the pressure between 
the journal and its bearing, and, consequently, increasing the 
friction. 

Law II states that, no matter how small the surface may 
be which presses against another, if the perpendicular pres- 
sure is the same, the friction will be the same. Therefore, 
large surfaces are used where possible; not to reduce the 
friction, but to reduce the wear and diminish the liability of 
heating. 

For instance, if the perpendicular pressure between a 
journal and its bearing is 10,000 pounds, and the coefficient 
of friction is .2, the amount of friction is 10,000 X .2 = 2,000 
pounds. 

Suppose that one-half the area of the surface of the jour- 
nal is 80 square inches; then, the amount of friction for 
each square inch of bearing is 2,000 ^ 80 = 25 pounds. 

If half the area of the surface had been 160 square inches, 
the friction would have been the same — that is, 2,000 pounds; 
but the friction per square inch would have been 2,000 -r- 
IGO = 12|- pounds, just one-half as much as before, and the 
wear and liability to heat would be one-half as great also. 



COEFFICIENTS OF FRICTION. 

TABLE 17. 



Description of Surfaces 
in Contact. 


Disposition 
of Fibers. 


state of the 
Surfaces. 


Coefficient 
of Friction. 


Oak on oak 


Parallel 
Parallel 
Parallel 
Parallel 
Parallel 
Parallel 


Dry 
Soaped 

Dry 
Soaped 

Dry 

Soaped 

Slightly 
Unctuous 

Slightly 
Unctuous 

Slightly 
Unctuous 


.48 


Oak on oak 


.16 


Wrought iron on oak 

Wrought iron on oak 

Cast iron on oak 


.62 
.21 
.49 


Cast iron on oak 


.19 


Wrought iron on cast iron. . 

Wrought iron on bronze 

Cast iron on cast iron 


.18 
.18 
.15 



ELEMENTARY MECHANICS 351 

EFFICIENCY. 

949. The force which is required to raise a weight, or 
overcome an equal resistance in any macliine, is always 
greater than this weight or resistance^ divided by the velocity 
ratio of the machine. 

Thus, if there were no friction, a machine whose velocity 
ratio was 5 would, by an application of 100 pounds of force, 
raise a weight of 500 pounds. 

Nqw, suppose that the friction in the machine is equiva- 
lent to 10 pounds of force, then it would take 110 pounds of 
force to raise 500 pounds. 

If, in the above illustration, friction were neglected, 110 
pounds X 5 = 550 pounds, or the weight that 110 pounds 
would raise; but, owing to the frictional resistance, it only 
raised 500 pounds; therefore, we have for the ratio between 

the two 4^ = .91. That is, 500 : 550 = .91 : 1. 
ooO 

950. This ratio between the weight actually raised and 
the applied force multiplied by the velocity ratio is called 
the efficiency of the macliiiie. 

Let /^= the force applied to the machine; 

F= the velocity ratio of the machine; 

W^= the weight actually lifted, or equivalent resistance 

overcome ; 

E = the efficiency of the machine. 

IV 
Then, E = ^^. (22.) 

Example. — In a machine having a combination of pulleys and gears, 
the velocity ratio of the whole is 9.75. A force of 250 pounds just lifts 
a weight of 1,626 pounds; what is the efficiency of the machine ? 

^ W 1,626 ^^^, n^r-^ . A 

SOLUTION.-E = -^ry = ^^Qy^.lh = '^^^^' '''" ^^ ^"' ' 

Since the total amount of friction varies with the load, it 

follows that the efficiency will also vary for different loads. 

Example. — A pulley block with a velocity ratio of 7 has an efficiency of 
88^ for a load of 3,000 pounds; what power is required to start the load? 

Solution.— E =-cTr^ ^^ -^^ = -h-n^ 

Hence. ^ = —^ =- = 1,128 pounds, nearly. Ans. 



352 ELEMENTARY MECHANICS. 

EXAMPLES FOR PRACTICE. 

1. A wedge is caused to move a weight vertically by means of a 
screw, which pulls the wedge horizontally on its base. If the screw 
has 5 threads per inch and the handle is 10 inches long, what force will 
be necessary to apply to the handle to raise a weight of 1,400 pounds, 
the height of the wedge being 8 inches, and the length, 14 inches ? 

" Ans. 2.546 lb., nearly. 

2. If the distance from the fulcrum to the point at which a force of 
^35 pounds is applied to a lever is 4 feet, and the distance from the 

• fulcrum to the weight is 1^ inches, how great a weight will the force 
lift? Ans. 4,3201b. 

3. It is desired to raise a weight of 600 pounds by means of a block 
and tackle having two fixed and two movable pulleys ; what force must 
be applied at the free end of the rope ? Ans. 150 lb. 

4. If, in the last example, the free end of the rope be attached to a 
windlass whose drum is 5 inches in diameter, and which has a handle 
situated 15 inches from the axis of the drum, what force will be neces- 
sary to raise the weight ? Ans. 25 lb. 

5. It is required to pull a wagon up an inclined plane one mile long. 
The height of the plane being 120 feet, and the weight of the wagon 
and load 2,816 pounds, what force will be necessary ? Ans. 64 lb. 

6. In Fig. 144, the radius of the wheel A is 32 inches, of C, 18 inches, 
and of E, 30 inches; the radius of the drum 7^ is 10 inches, of the pinion 
Z?, 4 inches, and of the pinion B, 6 inches. If P moves 4 feet 6 inches, 
how far will the weight W move ? Ans. f in. 

7. In the last example, how great a force must be applied at P to 
raise a weight of 2,160 pounds ? Ans. 30 lb. 

8. In example 4, if the friction be taken as 22^^ of the load lifted, 
what force will be necessary ? What will be the efficiency ? 

Ans. \ 30-5 lb. 

^, nearly. 



AVORK AND ENERGY. 

951. Work is the overcoming of resistance continually 
occurring along the path of motion. 

Mere motion is not work, but if a body in motion con- 
stantly overcomes a resistance, it does work. 

952. The unit of work is one pound raised vertically 
one foot^ and is called one foot-pound. All work is meas- 
ured by this standard. A horse going up hill does an 
amount of work equal to his own weight, plus the weight of 
the wagon and contents, plus the frictional resistances 



ELEMENTARY MECHANICS. 353 

reduced to an equivalent weight., multiplied by the vertical 
height of the hill. Thus, if the horse weighs 1,200 pounds, 
the wagon and contents 1,200 pounds, and the frictional re- 
sistances equal 400 pounds, then, if the vertical height of the 
hill is 100 feet, the work done is equal to (1,200 + 1,200 + 
400) X 100 = 280,000 foot-pounds. 

953. If the force necessary to overcome the resistance 
be represented by F^ the space through which the resistance 
acts by 5, and the work done by 6^, then U=- F S. 

If W=^ the weight of a body, and h = the height through 
which it is raised, [/= W h. Hence, the work done 
U=FS=Wh. (23.) 

954. The total amount of work is independent of time, 
whether it takes one minute or one year in which to do it ; 
but, in order to compare the work done by different machines 
with a common standard, time must be considered. 

When time is considered, the unit of time is always 07te 
minute^ and the unit which measures the capacity of any 
contrivance for producing work then becomes one foot- 
pound per minute. This unit is called the unit of 
poTver. The term power as here used has a different mean- 
ing from that previously given to it, which simply meant 
force acting to produce motion in simple machines. The 
student should carefully distinguish between the two. There 
will be no difficulty in doing this, as the wording of the 
sentence in which it occurs will always show whether force 
or work per minute is meant. The power of a machine 
may always be determined by dividing the work done in foot- 
pounds by the time in minutes required to do the work; i. e., 

FS 
Power =: -^, (24.) 

in which F and 5 have the same values as in formula 23, 
and 7^= the time in minutes. Hence, if a certain machine 
does, say, 10,000 foot-pounds of work in 10 minutes, its 
power is 10,000 ^ 10 = 1,000 foot-pounds per minute; if an- 
other machine does the same work in 5 minutes, its power 
is 10,000 — 5 = 2,000 foot-pounds per minute — iust twice as 



354 ELEMENTARY MECHANICS. 

much. Hence, we say that the power of the second ma- 
chine is twice that of the first, and understand thereby that 
if both machines work for the same length of time the 
second machine will do twice as much work as the first 
machine. 

955. Since the unit of power is very small, and would 
lead to the use of very large numbers in expressing the 
power of large machines, the common standard to which all 
work is reduced is the horsepower, which equals 33,000 units 
of power. 

One horsepower is 33^000 foot-pounds per minute; hi other 
words ^ it is 33^000 pounds raised vertically one foot in one 
minute^ or 1 pound raised vertically 33,000 feet in one minute, 
or any combination that will give 33,000 foot-pounds in one 
m,inute by multiplying the resistance in pounds by the distance 
in feet through which it is overcome^ and dividing by the time 
in minutes. 

Thus, 110 pounds raised vertically 5 feet in one second, is 
a horsepower; for, since one second = -gV of a minute, 110 X 
5 -f- gi^ r= 33,000 foot-pounds in one minute. The abbrevia- 
tion for horsepower is H. P. 

Example. — If the coefficient of friction is .3, how many horsepower 
will it require to draw a load of 10,000 pounds on a level surface a dis- 
tance of one mile in one hour ? 

Solution. — 10,000 X -3 = 3,000 pounds = the force necessary to over- 
come the resistance (resistance of the air is neglected). One mile = 5,280 
feet ; one hour = 60 minutes. 

_. FS 3,000 X 5,280 _, ^^^ . ^ , . ^ 

Power = „ = — ^^ = 264,000 foot-pounds per minute. 

y oO 

„ 264,000 „ T^ _, . 

Horsepower = -^^-tttjtt- = 8 H.P. Ans. 

956. Energy is a term used to express the ability of an 
agent to do work. Work cannot be done without motion, 
and the work that a moving body is capable of doing in being 
brought to rest is called the kinetic energy of the body. 

Kinetic energy means the actual energy of a body in 
motion. The work which a moving body is capable of doing 
in being brought to rest is exactly the same as the kinetic 



ELEMENTARY MECHANICS. 355 

energy developed by it when falling in a vacuum through a 

height sufficient to give it the same velocity. 

Let JV= the weight of the body in pounds; 

v = its velocity in feet per second ; 

A = the height in feet through which the body must 

fall to produce the velocity v; 

W 
m = the mass of the body = — . (See formula lOj 

Art. 888.) 

057. The work necessary to raise a body through a height 
h is W h. The velocity produced in falling through a height 



k\^ V = |/2^/j, and h = — — . (See formulas 15 and 16, 

'^£' 
Art. 896.) 

v^ W 

Therefore, work = Wh = W - — = 4- X — X v^ ^^mv^.ov 

^g g 

W/i = i7nv\ (25.) 

In other words^ if the weight of the body and its velocity are 
given, the work necessary to bring it to rest is equal to one- 
half the product of the mass and the square of tJie velocity in 
feet per second. This is the kinetic energy of a moving body. 

If a body, whose weight is 64.32 pounds, is moving with a 
velocity of 20 feet per second, the kinetic energy is -J mv"^ = 
^ 1 64.32 ^^, ,^^, 

%^''''= 23^^230 >< ^« = ^°° foot-pounds. 

958. The height through which a body would have to 
fall in a vacuum to gain a velocity of 20 feet per second is 
/ ^^' 400 400 ,. , , 

^^= 27= 2-^^^236= 64:32' '^^ "^^^ ^""^ ^^ ^"^^^^^ " 

body through this height is Wh = 64.32 X ttt^ = 400 foot- 

■ pounds, the same result as before. The distinction between 
kinetic energy and work, then, is this : A body moving with 
a certain velocity^ and meeting with no resistance, is not doing 
any work, but has a certain amount of kinetic energy stored 
up in it, which depends upon the velocity and mass of the 
body. 



356 ELEMENTARY MECHANICS. 

If a resistance be interposed just sufficient to stop the 
body, the body will then do an amount of work equal to its 
kinetic energy. 

Example. — If a body weighing 25 pounds falls from a height of 100 
feet, how much work can be done ? 

Solution.— Work = Wh = 25 X 100 = 2,500 foot-pounds. Ans. 

Example. — A body weighing 50 pounds has a velocity of 100 feet per 

second ; what is its kinetic energy ? 

T^- .' , , ^^^ 50 X 1002 „ „^^ ^^ 
Solution. — Kmetic energy = \m v^ = -^ = 2 v S2 1 fi ~ 7,773.63 

foot-pounds. Ans. 

Example. — In the last example, how many horsepower would be 
required to give the body this amount of kinetic energy in 3 seconds ? 

Solution. — 1 H.P. =33,000 pounds raised one foot in one minute. 

If 7,773.63 foot-pounds of work are done in 3 seconds, in one second 

7 77'-? 63 
there would be done ' „ ' — = 2,591.21 foot-pounds of work, and in 

o 

one minute, 2,591.21 X 60 = 155,472.6 foot-pounds. 

The number of horsepower developed would be 

959. Potential energy is latent energy ; it is the energy 

which a body at rest is capable of giving out under certain 
conditions. 

If a stone be suspended by a string from a high tower, it 
has potential energy due to its position. If the string be 
cut, the stone will fall to the ground, and during its fall its 
potential energy will change into kinetic energy, so that, at 
the instant it strikes the ground, its potential energy is 
wholly changed into kinetic energy. 

At a point equal to one-half the height of the fall, the 
potential and kinetic energies were equal. At the end of 
the first quarter, the potential energy was J, and the kinetic 
energy \\ at the end of the third quarter, the potential 
energy was ^, and the kinetic energy |. 

A pound of coal has a certain amount of potential energy. 
When the coal is burned, the potential energy is liberated, 
and changed into kinetic energy in the form of heat. The 
kinetic energy of the heat changes water into steam, which 



ELEMENTARY MECHANICS. 357 

thus has a certain amount of potential energy. The steam 
acting on the piston of an engine causes it to move through 
a certain space, thus overcoming a resistance, changing the 
potential energy of the steam into kinetic energy, and thus 
doing work. 

Potential energy , then, is the energy stored within a body, 
which may be liberated and produce motion, thus generating 
kinetic energy, and enabling work to be done. 

960. The principle of conservation of energy teaches 
that energy, like matter, can never be destroyed. If a clock 
is put in motion, the potential energy of the spring is changed 
into the kinetic energy of motion, which turns the wheels, 
thus producing friction. The friction produces heat, which 
dissipates into the surrounding air, but still the energy is 
not destroyed — it merely exists in another form. The poten- 
tial energy in coal was received from the sun, in the form of 
heat, ages ago, and has laid dormant for millions of years. 



FORCE OF A BLOW. 

961. The average force of a blo^w may be determined 
as follows : 

In driving a nail into a piece of wood with a hammer, the 
head of the hammer must be capable of exerting, in a very 
short time, a force equal to that of a load sufficiently heavy 
to produce by its weight a movement of the nail into the wood, 
equal to the movement of the nail produced by the hammer; 
in other words, tJie striking force, multiplied by the distance 
that the nail is driven into the wood, must equal the kinetic 
energy of the hammer. 

Suppose that the velocity of the hammer, as it strikes the 
nail, is 30 feet per second, that the weight of the head is 2 
pounds, and that the nail is driven into the wood \ of an 
inch. Let the resistance offered by the wood, which is the 
same as the striking force of the hammer, be represented by 
F; then, since \ inch = ^ of a foot, 

^X A = i^^^ X -c^' = ^^3^.^g X 30' = 28, nearly. 
Therefore, F= 28 X 48 = 1,344 pounds. 
D. 0. lll.^Q 



358 ELEMENTARY MECHANICS. 

Had the penetration been ^ of an inch, instead of ^ of an 
inch, the value of i^ would have been twice as large, or 2,688 
pounds. This is as it should be, since the work done is the 
same in both cases, while the distance through which the 
force acts is only half as great in the second case as in the 
first case. Hence, in order that the work may be the same, 
the force must be doubled. 

Example. — If the head of a drop-hammer, weighing 400 pounds, falls 
from a height of 10 feet, and compresses a piece of cold iron .01 of an 
inch, what was the striking force of the hammer ? 

Solution. — .01 of an inch = --^^oi a foot. 

j^X ■-U-= fF^ = 400 X 10 =*: 4,000. 

Hence, 7^= ^^^5^^^ = 4,800,000 lb. Ans. 



DENSITY AND SPECIFIC GRAVITY. 

962. The density of a body is its mass divided by 

its volume in cubic feet.* 

Let I? be the density; then, the density of a body is, 

m WW 

D — -YJ. Since 7;^ = — , i> = — f>. (26.) 

Example. — Three cubic feet of cast iron weigh 1,350 pounds; what 
is the density of cast iron ? 

„ W 1,350 _.„ 

SoLUTiON.-i^ = 7F = 32:16^ = ^^•^^^- 

Example. — A cubic foot of water weighs 63.43 pounds; («) what is 
its density ? {b) What is the ratio between the density of cast iron and 
the density of water ? 

Solution. — id) D = — rr = -kk — ^ — r = 1-941 = density. Ans. 
^ ^ gV 33.16 X 1 

{^) -i f^A-, = 7.31 = ratio. Ans. 
1.941 

963. The specific gravity of a body ts the ratio 
between its weight and the weight of a like volume of water. 

*NoTE. — Some writers define density as the weight of a unit of 
volume of the material. When English measures are used, the density 
of any material, according to this definition, is the weight of a cubic 
foot of the material in pounds. 



ELEMENTARY MECHANICS. 359 

964. Since gases are so much lighter than water, it is 
usual to take the specific gravity of a gas as the ratio 
between the weight of a certain volume of the gas and the 
weight of the same volume of air. 

Example. — A cubic foot of cast iron weighs 450 pounds; what is its 
specific gravity, a cubic foot of water weighing 62.42 pounds? 

Solution. — According to the definition, 

450 



62.42 



= 7.21. Ans. 



Note. — Notice that this answer is the same as that of the preceding 
example ; hence, the specific gravity of a body is also the ratio of the 
density of a body to the density of water. 

965. The specific gravities of different bodies are given 
in printed tables; hence, if it is desired to know the weight 
of a body that cannot be conveniently weighed, calculate its 
cubical contents^ and multiply the specific gravity of the body 
by the weight of a like volume of water ^ remembering that a 
cubic foot of water weighs 62. Ji2 pounds. 

Example. — How much will 3,214 cubic inches of cast iron weigh? 
Take its specific gravity as 7.21. 

Solution. — Since 1 cubic foot of water weighs 62.42 pounds, 3,314 
cubic inches weigh 

rS§ ^ ^^•'^^ ^ 116.098 pounds. 
Then, 116.098 X 7.21 = 837.067 pounds. Ans. 

Example. — What is the weight of a cubic inch of cast iron ? 

62.42 
Solution. — ^ x 7.21 = .26044 pound. Ans. 
1 , < <&o 

One cubic foot of pure distilled water at a temperature of 
39.2° Fahrenheit weighs 62.425 pounds, but the value usually 
taken in making calculations is 62.5 pounds. 

Example. — What is the weight in pounds of 7 cubic feet of oxygen ? 

Solution. — One cubic foot of air weighs .08073 pound, and the 
specific gravity of oxygen is 1.1056, compared with air; hence, 
.08073 X 1. 1056 X 7 = . 62479 pound, nearly. Ans. 



EXAMPLES FOR PRACTICE. 

1. A man jumps from a railroad train travehng at the rate of 60 
miles per hour; if he weighs 160 lb., what is the kinetic energy of his 
body ? , Ans. 19,263.68 ft. -lb. 



360 ELEMENTARY MECHANICS. 

2. A hammer strikes a nail with a velocity of 40 ft, per sec. ; if the 
head weighs 1^ lb. , and the nail is driven -^^ in. , what is the force of the 
blow? Ans. 1,433 lb., nearly. 

3. A load of 20,000 lb. is pulled up an inclined plane 1| miles long 
in 5 minutes ; if the height of the plane is 800 ft., and the force acting 
along the plane necessary to overcome friction is 8^ of the load, what 
is the horsepower required ? Ans. 125.77 H. P. 

4. If a cubic foot of a certain substance weighs 162^ lb. , what is its 
specific gravity ? Ans. 2. 6. 

5. A mixture of lead and tin has a specific gravity of 9.28; what is 
the weight of a cubic inch ? Ans. 5.37 oz. 

6. A weight of 11,625 lb. is raised vertically 10 ft. in 3 minutes, by 
means of a block and tackle and a windlass ; the f rictional resistances 
being 26^, how many horsepower were required ? Ans. 1.48 H. P. 

7. A body weighing 24,062.5 lb. is drawn on a horizontal surface at 
the rate of 600 ft. per min. ; the coefficient of friction being 8%, what 
horsepower will be necessary to overcome the resistance of friction ? 

Ans. 35 H. P. 

8. A solid cast iron sphere, 12" in diameter, falls through a height 
of 50 ft. ; what will be its kinetic energy on striking ? Ans. 11,781 ft.-lb. 

966. The table of specific gravities gives the specific 
gravities of a variety of substances likely to be met with in 
ordinary practice. The weights per cubic foot are calcu 
lated on a basis of 62. 5 pounds of water per cubic foot. 



STRENGTH OF MATERIALS. 



MATERIALS USED FOR CON- 
STRUCTION. 

1331. The principal materials used in engineering con- 
struction are timber ^ brick, stone, cast iron, wrought iron, and 
steel. Table 23 gives their average weights per cubic foot. 

TABLE 23. 





Average W 


sight. 


Approximate 
Weight of Piece 
1" Square and V 

Long in Lb. 


Material. 


Pounds per 
Foot. 


Cubic 


Timber 


40 
125 
160 
450 
480 
490 


.278 


Brick 




Stone 




Cast Iron 


3.125 


Wrought Iron 


3.333^ 
3.403 


Steel 







CAST IRON. 

1 332. Cast iron is a combination of pure iron with 
from 2^ to 6^ of carbon. 

Pig iron is the result of the first smelting, and is obtained 
directly from the ore. Pig iron is rarely, if ever, used for 
anything except to be remelted and made into cast iron or 
wrought iron. 

1333. Cast iron is of two kinds, white cast iron and 
gray cast iron. The first is a chemical compound of iron 
with from 2^ to 6^ of carbon, nearly all the carbon being 
chemically combined with the iron. The second, or gray 
cast iron, contains a part of the carbon in chemical combi- 
nation, and the rest in the state of graphite mechanically 
mixed with the iron. When a piece of gray cast iron is 

For notice of the copyright, see page immediately following the title page. 



742 STRENGTH OF MATERIALS. 

broken, a large number of black specks are seen on the 
broken ends ; these specks are pure carbon in the form of 
graphite. 

1334, "White cast iron contains hardly any free car- 
bon. It is of two kinds, granular and crystalline. Both 
are very hard and brittle, and are only used for conversion 
into wrought iron or steel. 

1335» Gray cast iron is divided into three classes, 
known as Nos. 1, 2, and 3. 

No. 1 contains the largest amount of carbon in mechani- 
cal mixture, which makes it soft and fusible, though not as 
strong as Nos. 2 and 3. It is very suitable for making cast- 
ings where precision in form is required, as, owing to its 
fluidity, it fills the mold well; but it is not suitable for 
castings requiring strength. 

No. 2 is most suitable for use in constructions, as it is 
stronger than No. 1, and not so soft. 

No. 3 contains the smallest amount of carbon in the 
graphitic (uncombined) form, and is, in consequence, harder 
and more brittle. It is fit only for the massive and heavy 
parts of machinery. 

1336. Cast iron has certain advantages and disadvan- 
tages as a material for engineering construction. It is easy 
to give it any desired form. A pattern of the piece desired 
is made ; a mold is made in the sand, the pattern is removed, 
and the melted iron poured in. Cast iron resists oxidation 
(rust) better than either wrought iron or steel. Its com- 
pressive (crushing) strength is very high, but its tensile 
(stretching) strength is comparatively low. It cannot be 
riveted, or welded by forging. It is brittle, breaking off 
without giving much warning, and stretching but little be- 
fore giving away. It is liable to have hidden and small 
surface defects and air bubbles, and this makes its strength 
uncertain. 

Another serious drawback in the use of cast iron is its 
liability to initial stresses from inequality in cooling after 
having been poured into the molds. Thus, if one part of 



STRENGTH OF MATERIALS. 743 

the casting is very thin and another very thick, the thin 
part cools first, and, in cooling, contracts; the thick part, 
cooling afterwards, causes stresses in the thin part, which 
may be sufficient to break it, or, if not, there may be so 
great a stress in the thin part that a small additional force 
will break it. 

Cast iron is not well adapted to a tensile stress, nor to re- 
sist shocks. It is used for columns and posts in buildings, 
on account of its high compressive strength. In machinery, 
it is used in all those parts where weight, mass, or form is of 
more importance than strength, as in frames and bed plates 
of machines, and for hangers, pulleys, gear wheels, etc. It 
is also used for water mains where the pressure to be 
resisted is not too great. 

ISST. Malleable cast iron is made by heating the 
casting in an annealing oven, in powdered hematite ore. It 
can be hammered into any desired shape when cold, but is 
very brittle when hot. 

^WROUGHT IRON. 

1338. Wrought iron is the product resulting from 
the reduction of the carbon in cast iron. It is obtained by 
melting white cast iron, and passing an oxidizing flame over 
it. When the carbon is burned out, the mass of iron is left 
in a pasty condition, full of holes. It is then taken out, and 
hammered or rolled in order to unite it into one mass. The 
result of this first process is not suitable for use in any con- 
struction of importance, and has to be reheated and rerolled 
a number of times, in order to make it homogeneous, and to 
remove flaws from within the iron. 

At best, therefore, wrought iron is a series of welds ; if a 
piece is broken, the separate layers of which it is composed 
can be seen plainly. It cannot be melted and run into 
molds, as can be done with cast iron ; but it can be easily 
welded by forging; that is, two pieces of wrought iron can 
be united by raising them to the proper temperature and 
hammering them together. Wrought iron is much more 
capable of bearing a tensile or transverse stress than cast 



744 STRENGTH OF MATERIALS. 

iron; it is tougher, stretches more, and gives more warning 
before fracture. 

It withstands shocks far better than cast iron. Two 
pieces may be punched or drilled and riveted together. The 
entire process weakens the iron, and cast iron would not 
withstand it. It has also to withstand flanging and the 
stresses due to changes of temperature. 

Wrought iron cannot be hardened like steel by heatmg 
and then dipping in water, but may be case-hardened by 
rubbing the surface with potassium cyanide or potassium 
ferrocyanide while at a cherry-red heat and then dipping in 
water. The cyanide causes the iron to be carbonized to a 
slight depth ; that is, through a depth of about -^-^ of an inch 
the iron is converted into steel which can be hardened. 
Cast iron may be hardened in the same way. 

The quality of wrought iron varies considerably, and the 
terms by which it is known in the market refer to the 
amount of working which the iron has received. We thus 
have common bar iron, best iron, double best, and triple best. 
These terms are only rough indications of quality. When 
wrought iron is rolled cold under great pressure, it has a 
smooth polished surface, and its strength is greatly increased. 

When the word iron is used alone, wrought iron is meant. 



STEEL. 



1 339. Steel is a chemical compound of iron and carbon ; 
that is, it contains no carbon in a free state, as cast iron does. 
Its tensile strength is greater than that of wrought iron, 
and its compressive strength greater than that of cast 
iron. It is by far the strongest material used in the me- 
chanic arts. Its strength varies greatly with its purity and 
the amount of carbon it contains. According to the amount 
of carbon in it, steel is divided into high grade, medium 
grade, and low grade, the high grades having the most carbon. 
Steel, unlike wrought iron, is fusible; unlike cast iron, it 
can be forged ; and, with the exception of the higher grades, 



STRENGTH OF MATERIALS. 



745 



It can be welded by heating and hammering, although care 
must be exercised in so doing. 

1340. The special characteristic of steel (the very low- 
est grades excepted) is that, when it is raised to a cherry 
red heat and suddenly cooled, it becomes brittle and exceed- 
ingly hard, and that by subsequent heating and slow cooling, 
the hardness may be reduced to any desired degree down to 
the point of least hardness that steel possessing that amount 
of carbon can have. The first process is called Jiardcning^ 
and the second tempering. 

If the surface of a piece of steel that has been hardened is 
polished slightly so as to remove the dark scale or soot which 
covers it, and is then reheated, it will be found that, as the 
temperature increases, a series of colors appear. These 
colors are always the same for the same temperature, and, 
if the steel is suddenly cooled when one of the colors ap- 
pears, it acquires a degree of hardness which is always the 
same for the same color and for the same quality of steel. 

In Table 24 are given the colors with the corresponding 
temperatures that occur in tempering different kinds of tools. 

TABLE 24. 



Tools. 


Color. 


Temperature, 
Fahr. 


Lancets 


Pale yellow 
Pale straw 

Yellow 

Brown 

Red 

Purple 

Blue, bright 

Blue, full 

Blue, dark 


430° 


Razors and scalpels 


450° 


Penknives, chisels for cast 
iron, and screw taps 

Scissors and chisels for 
wrought iron 


470° 
490° 


For carpenters' tools in gen- 
eral 

Fine watch springs and table 
knives 


510° 
530° 


Swords and lock springs 

Daggers, fine saws and needles 
Common saws and springs. . . 


550° 
560° 
G00° 



746 STRENGTH OF MATERIALS. 

Steel that has been hardened may be softened by heating 
it to a hardening temperature and then cooHng it very 
slowly; this process is called annealing. 

1341. Steel is made in one of the three following ways: 

1. By adding carbon to Avrought iron. 

2. By removing carbon from cast iron. 

3. By melting together cast and wrought iron in suitable 
proportions. 

Several processes, varying with the quality of the product 
desired and the use for which it is intended, are used in 
making steel. The processes in general use are the 
following : 

(a) Cementation, in which bars of very pure wrought 
iron are heated to a high temperature in contact with car- 
bon. The product, known as blister steel, is used for 
cutlery, tools, etc. 

(l?) Crucible steel, also called cast steel, is made by 
melting pure wrought iron in a crucible with enough char- 
coal and cast iron to introduce the required amount of 
carbon. It is used for making springs, cutlery, tools, etc. 

(c) Dessemer steel is made by decarbonizing cast iron 
by forcing a powerful blast of air through a melted mass of 
the iron. This removes the greater part of its carbon. A 
small quantity of very pure cast iron, rich in carbon, is 
then added, bringing up the percentage of carbon to the 
required amount. 

(d) Open-hearth, steel is made by fusing a charge 
consisting of the suitable proportions of cast iron with 
wrought iron scrap, or with Bessemer steel scrap. 

134:2>» Bessemer and open-hearth steel contain more 
impurities than blister and crucible steel do; but they are 
much cheaper, and are just as suitable for many purposes. 
It is only in consequence of the introduction of these two 
cheap varieties that steel can be extensively used, as blister 
and crucible steels would, in the majority of cases, be too 
expensive. 



STRENGTH OF MATERIALS. 747 

Bessemer and open-hearth steels contain from .05 fo to 1-|-^ 
of carbon. The proportion of carbon in the best kinds of 
tool and cutlery steels is as follows: 

Razor steel, l^fo. Very difficult to forge, and easily burnt. 
Saw file, If^. Bears heat not above cherry redness. 

Tool steel, l^fo. Ordinary cutting tools. Welds with 

difficulty. 
** " 1^^. For mandrils and heavy cutting tools. 

** ** 1^. For chisels, gravers, etc. 

At the Imperial Works at Neuberg, Austria, the follow- 
ing percentages of carbon are present in the different 
grades of Bessemer steel: 

1.58^ to 1.38^. Cannot be welded, and is rarely used. 
1.38^ " 1.12^. Great care must be used in working. 
1.12^ " .88^. Welds easily; used for bits, chisels, etc. 
.88^ *' .62 fo. Used for cutting tools, files, etc. 
.62 fo " .38^. Mild steel ; for tires, etc. 
.38^ " .15^. Tempers slightly; for boiler plates and 

axles. 
.15^ " .05^. Does not temper; steel for pieces of 

machinery. 

1343. It will be noticed from what precedes that the 
hardness of steel depends upon the amount of carbon it 
contains. 

Some kinds of crucible cast steel can be hardened by heat- 
ing to a low red heat and then allowing them to cool slowly 
in the air without dipping in water. They are called self- 
hardening steels, the best known being Mushet's special tool 
steel. This contains about 2^ carbon with 7^ to 12^ tung- 
sten in alloy with the iron. The same property is character- 
istic of Hadfield's manganese steel, which contains between 
.8^ and 1.2;^ of carbon and 7^ to 20^ of manganese. 

As a rule, when a piece of steel is broken across the grain, 
the finer the grain and the whiter and cleaner the fracture 
the more carbon it contains. 



748 STRENGTH OF MATERIALS. 

STRESSES AND STRAINS. 

1344. The molecules of a solid or rigid body being held 
together by the force of cohesion, this force must be over- 
come to a greater or less degree in order to change the form 
and size of the body, or to break it into parts. The internal 
resistance which a body offers to any force tending to over- 
come the force of cohesion is called a stress. If a weight 
of 1,000 pounds is held in suspension by a rod, there will be 
a stress of 1,000 pounds in the rod. In this country and 
England, stresses are measured in pounds or tons ; in nearly 
all other civilized countries, in kilograms. Whenever a body 
is subjected to a stress, the total stress induced by the act- 
ing force at any section of the body is the same as the total 
stress at any other section. 

1 345. The unit stress (called also the intensity of 

stress) is the stress per unit of area; or, it is the total stress 
divided by the area of the cross-section. In the above illus- 
tration, if the area had been 4 sq. in. the unit stress would 

have been -^- — = 250 lb. per sq. in. Had the area been 

— sq. in., the unit stress would have been — ^-^ — = 2,000 lb. 

per sq. in. 

Let P= the total stress in pounds; 

A = area of cross-section in square inches; 
S = unit stress in pounds per square inch. 

Then, S = ^,orP=A S. (108.) 

That is, ^/le total stress equals the area of the section^ inulti- 
plied by tJie unit stress. 

When a body is stretched, shortened, or in any way 
deformed through the action of a force, the amount of defor- 
mation is called a strain. Thus, if the rod before mentioned 
had been elongated yV by the load of 1,000 pounds, the 
strain would have been ^y. Within certain limits, to be 
given hereafter, strains are proportional to the stresses pro- 
ducing them. 



STRENGTH OF MATERIALS. 749 

1346. The unit strain is the strain per unit of length 
or of area, but is usually taken per unit of length and called 
the elongation per unit of length. In this paper, the unit 
of length will be considered as one inch. The unit strain, 
then, equals the total strain divided by the length of the 
body in inches. 

Let/=: length of body in inches; 
e ■= elongation in inches; 
s = unit strain. 

Then, s~- ox e^ls. (109.) 

1347. Whenever a force, no matter how small, acts 
upon a body, it produces a stress and a corresponding strain. 

According to the manner in which forces act upon a 
body, the stresses are divided into the following classes: 

1. Tension^ which produces a tensile or pulling stress. 

2. Compression^ which produces a compressive or crush- 
ing stress. 

3. Shear ^ which produces a shearing or cutting stress. 

4. Torsion^ which produces a torsional or twisting stress. 

5. Flexure y which produces a transverse or bending stress. 



TENSION. 

1348. When two forces act upon a body in opposite 
directions (away from each other) the body is said to be in 
tension.' The two forces tend to elongate the body and thus 
produce a tensile stress and strain. A weight supported by 
a rope affords a good example. The weight acts downwards, 
and the reaction of the support to which the upper end of 
the rope is fastened acts upwards; the result being that the 
rope is stretched more or less, and a tensile stress is produced 
in it. Another familiar example is the connecting rod of a 
steam engine on the return stroke. The cross-head then 
exerts a pull on one end of the rod, which is resisted by the 
crank-pin on the ether. 



750 STRENGTH OF MATERIALS. 

Example. — An iron rod, 2 inches in diameter, sustains a load of 
90,000 pounds; what is the unit stress ? 

Solution. — Using formula 108, 

P = AS,ov S=~ = J^;^^l^, = 28,647.82 lb. per sq. in. Ans. 
A 2^ X. 7854 ^ ^ 



EXPERIMENTAL LAWS. 

1349. The following laws have been established by 
experiment: 

1. When a body is subjected to a small stress^ a small strain 
is produced^ and when the stress is removed the body springs 
back to its original shape. This leads to the conclusion that^ 
for small stresses^ bodies are perfectly elastic. 

2. Within certain limits, the change of shape (strain) is 
directly proportional to the applied force. 

3. WJien the stress is sufficiently great, a strain is pro- 
duced wJiich is partly permanent ; that is, the body does not 
spring back entirely to its original form when the stress is 
removed. This lasting part of the strain is called a set, and 
in such cases the strain is not proportional to the stress. 

4. Under a still greater stress, the strain rapidly increases, 
and the body is finally ruptured or broken. 

5. A force acting suddenly, as a shock, causes greater 
injury tJian a force gradually applied. 

According to the first law, the body will resume its 
original form when the force is removed, provided the stress 
is not too great. This property is called elasticity. Accord- 
ing to the second law, the strain is proportional to the stress 
within certain limits. Thus, if a pull of 1,000 pounds 
elongates a body .\" , a pull of 2,000 pounds would elongate 
it .2". This is true up to a certain limit, beyond which the 
body will not resume its original form upon the removal of 
the stress, but will be permanently strained more or less, 
according to the amount of stress. The stress at the point 
where the set begins is called the elastic limit. All 
strains produced by stresses within (less than) the elastic 
limit are directly proportional to the stresses. 



STRENGTH OF MATERIALS. 751 

1350. Stresses are equal but opposed to the external 
forces producing them, and are, therefore, measured and 
represented by these forces. Thus, as we have explained 
before, a force of 1,000 pounds produces a stress of 1,000 
pounds. The external force is the force applied to a fixed 
body; the stress is the resistance offered by the body to a 
change of form; and when the body ceases to change (as 
when a rod ceases to elongate), the stress just balances the 
external force. 

COEFFICIENT OF ELASTICITY. 

1351. Amongst engineers, the term elasticity means 
the resistance which a body offers to a permanent change of 
form ; and by strength^ the resistance which a body offers to 
division or separation into parts. 

It follows from this that those bodies which have the 
highest elastic limit are the most elastic. 

1352. The coefficient of elasticity is the ratio of 

the unit stress to the unit strain, provided the elastic limit 
is not exceeded. Let 5 be the unit stress, s the unit strain, 
and E the coefficient of elasticity; then, by definition, 

5 . . 

^ = — . Substituting the values of ^ and s obtained from 

"formulas 108 and 109, 

1353. If in this formula we assume ^ = /, and A = l 
(1 square inch), then E =^ P. That is, the coefficient of 
elasticity is that force which, if stress and strain continued 
proportional to each other, would produce in a bar of unit 
area a strain equal to the original length of the bar (/= e). 
This, however, is never the case, as the elastic limit and the 
ultimate strength are reached before the applied force 
reaches the value E. 

Example. — A wrought iron bar 2 inches square and 10 feet long is 
stretched .0528 inch by a stress of 44,000 pounds; what is the coeffi- 
cient of elasticity ? 

Solution. — Using formula 1 lO, 

^ PI 44,000X10X12 o^nnnnnrMK • a 

E--^^= — 2^ X .0528 ~ 25.000.000 lb. per sq. in. Ans. 



752 



STRENGTH OF MATERIALS. 



1354. The ultimate strength of any material is that 
unit stress which is just sufficient to brealc it. 

1355. The ultimate elongation is the total elonga- 
tion produced in a unit of length of the material having a 
unit of area, by a stress equal to the ultimate strength of 
the material. 

1356. For the same size, quality, and kind of material, 
the ultimate strength, ultimate elongation, coefficient of 
elasticity, and elastic limit are the same for different pieces. 
Table 25 gives the average values of the coefficient of 
elasticity (^J, elastic limit (i^,), ultimate strength (5J, and 

TABLE 25. 





Coefficient 


Elastic 


Ultimate 


Ultimate 


Material. 


of Elasticity. 


Limit. 


Strength. 


Elongation. 




Lb. per Sq. In. 


Lb. per Sq. In. 


Lb. per Sq. In. 


In. per Linear 
Inch. 


Timber 


1,500,000 


3,000 


10,000 


0.015 


Cast Iron .... 


15,000,000 


6,000 


20,000 


0.005 


Wrought Iron 


25,000,000 


25,000 


55,000 


0,20 


Steel 


30,000,000 


50,000 


100,000 


0.10 



ultimate elongation (^'J, of different materials, the quantities 
given being for tension only. As brick and stone are never 
used in tension, their values are not given. 

The values in this table are subject to great variation, and 
cannot be depended upon in designing machine parts. Thus, 
the ultimate tensile strength of steel varies from less than 
60,000 to more than 180,000 pounds per square inch, accord- 
ing to its purity and the amount of carbon it contains; that 
of cast iron from 12,000, or 13,000, to over 40,000; wrought 
iron varies from 40,000 to 72,000, according to quality, the 
latter value being for iron wire. Timber varies fully as 
much as, if not more than, any of the three preceding 
materials, its properties depending upon the kind of wood^ 
its degree of dryness, the manner of drving, etc. 



STRENGTH OF MATERIALS. 753 

All the problems in this section will be solved by using the 
average values given in the preceding and following tables; 
the designer, however, should not use them, but either test 
the materials himself or state in the specifications what 
strength the material must have. For example, mild steel, 
for boiler shells, should have a tensile strength of not less 
than 60,000 or 65,000 pounds per square inch; Bessemer 
steel, for steel rails, not less than 110,000; open-hearth steel, 
for locomotive tires, not less than 125,000, and crucible cast 
steel, for tools, cutlery, etc., not less than 150,000. It is 
also customary to specify the amount of elongation. This is 
necessary because, as a rule, the elongation decreases as the 
tensile strength increases. Having tested the material 
about to be used, or having specified the lowest limits, the 
designer can ascertain the strength and stiffness of construc- 
tion by means of the formulas and rules which are to follow. 

Example. — How much will a piece of steel 1 inch in diameter and 
1 foot long elongate under a steady load of 15,000 pounds ? 

Solution. — Ex = -—: — , or ^ = 



From Table 25, £1 = 30,000,000 for steel; hence, 

- 15,000 X 12 _ f.(.r^n.„ A 

^ ~ 12 X .7854 X 30,000,000 ~ * ^"^' 

Note. — All lengths given in this treatise on Strength of Materials 
must be reduced to inches before substituting in the formulas. 

Example. — A piece of timber has a cross-section 2" X 4" and is 
6 feet long. A certain stress produces an elongation of .144 inch; 
what is the value of the stress in pounds ? 

Solution. — 

E,=^^,orP = ^ = '''''^'^'\><^^^^><'''^ = 2^,000lh. Ans. 



COMPRESSION. 

135T, If the length of the piece is not more than five 
times its least transverse dimension (its diameter, when 
round; its shorter side, when rectangular, etc.), the laws of 
compression are similar to those of tension. The strain is 
proportional to the stress until the elastic limit has been 

D. o. ni.—u 



754 



STRENGTH OF MATERIALS. 



reached ; after that, it increases more rapidly than the stress, 
as in the case of tension. The area of the cross-section is 
slightly enlarged under compression. In Table 26 are given 
the average compression values of E^ Z, and S for wood, 
brick, stone, cast iron, wrought iron, and steel. (See also 
Table 25.) E is not given for brick, nor L for cast iron, 
brick, or stone, because these values are not known. To 
distinguish between tension and compression when apply- 
ing a formula, Z^, Z^, and S^ will be used instead of E^^ Z^, 



and 5j, 



TABLE 26. 



Material. 


Coefficient of 

Elasticity. 

Z.. 


Elastic 
Limit. 


Ultimate 

Compressive 

Strength. 

5.. 


Timber 


Lb. per Sq. In. 

1,500,000 


Lb. per Sq. In. 

3,000 

25,000 
50,000 


Lb. per Sq. In. 

8,000 
2,500 


Brick 


Stone 


6,000,000 
15,000,000 
25,000,000 
30,000,000 


6,000 

90,000 

55,000 

150,000 


Cast Iron 


Wrought Iron 

Steel 





1358. When the length of a piece subjected to com- 
pression is greater than ten times its least transverse dimen- 
sion, it is called a column^ and the material fails by a side- 
wise bending or flexure. The preceding table is to be used 
only for pieces whose length does not exceed five times the 
least dimension of the cross-section. (See Art. 1421<3:.) 

Example. — How much will a wrought iron bar 4 inches square and 
15 inches long shorten under a load of 100,000 pounds ? 

Solution. — Za = ^— , or ^ = „ . 



Hence, e = 



100,000 X 15 
16 X 25,000.000 



= .00375". Ans- 



STRENGTH OF MATERIALS. 



755 



SHEAR. 
1359. When two surfaces move in opposite directions 
very near together in such a manner as to cut a piece of 
material, or to pull part of a piece through the remainder, 
the piece is said to be sheared. A good example of a shear- 
ing stress is a punch; the two surfaces in this case are the 
bottom of the punch and the top of the die. Another ex- 
ample is a bolt with a thin head ; if the pull on the bolt is 
great enough, it will be pulled through the head and leave a 
hole in it, instead of the bolt breaking by pulling apart, as 
would be the case with a thick head. In this case, the two 
surfaces are the under side of the head and the surface 
pressed against. Other examples are a knife cutting a piece 
of wood, and the ordinary shears from which this kind of 
stress takes its name. 

TABLE 27. 



Material. 



Timber (across the grain) 
Timber (with the grain) . 

Cast Iron 

Wrought Iron •. . 

Steel 



Coefficient of 
Elasticity. 



400,000 

6,000,000 

15,000,000 



Ultimate 
Shearing 
Strength. 



3,000 

600 

20,000 

50,000 

70,000 



1360. Formula 108 applies in cases of shearing stress, 
but formulas 109 and llO are never used for shearing. 
In the preceding table, E^ and S^ are used to represent, 
respectively, the coefficient of shearing elasticity, and 
ultimate shearing strength. 

Example. — What force is necessary to punch a one-inch hole in a 
wrought iron plate -f of an inch thick ? 

Solution. — 1" x 3.1416 X -f" = 1.1781 sq. in. = area of punched sur- 
face = area of a cylinder 1" in diameter and -f" high. Using 

formula 108, 

P = AS^ = \. 1781 X 50,000 = 58,905 lb. Ans. 



756 STRENGTH OP MATERIALS. 

Example. — A wooden rod 4 inches in diameter and 2 feet long is 
turned down to 3 inches diameter in the middle so as to leave the 
enlarged ends each 6 inches long. Will a steady stress pull the rod 
apart in the middle, or shear the ends ? 

Solution.— P = ^ ^3 = 3 X 3.1416 X 6 X 600 = 23,620 lb. to shear off 
the ends. 

The force required to rupture by tension is 

F = A Si=2'X .'7854 X 10,000 = 31,416 lb. 

Since the former is only about f of the latter, the piece will fail 
through the shearing off of the end. Ans. 

Had a transverse stress been used, the force necessary to 
shear off a section of the end would have been 

4' X .7854 X 3,000 = 37,700 lb. 



FACTORS OF SAFETY. 

1361. It was previously stated that no stress should 
ever be applied to a machine part that would strain it 
beyond the elastic limit. The usual practice is to divide 
the ultimate strength of the material by some number 
depending upon the kind and quality of the material, and 
upon the nature of the stress ; this number is called a factor 
of safety. 

TAe factor of safety for any material is the ratio of its 
ultimate strength to the actual stress to which it is subjected^ 
or for which it is intended. 

In Table 27, 70,000 pounds per square inch is given as 
the ultimate shearing strength for steel. Now, suppose that 
the actual stress on a piece of steel is 10,000 pounds per 
square inch ; then, the factor of safety for this piece would 
, 70,000 

^"io:w=^- 

1362. To find the proper allowable working strength 
of a material, divide the ultimate strength for tension, com- 
pression, or shearing, as the case may be, by the proper 
factor of safety. 

Table 28 gives the factors of safety generally used in 
American practice. Factors of safety will always be 
denoted by the letter y in the formulas to follow. 



STRENGTH OF MATERIALS. 



757 





TABLE 


28. 




Material. 


For Steady 

Stress. 
(Buildings.) 


For Varying 

Stress. 
(Bridges.) 


For Shocks. 
(Machines.) 


Timber 


8 

15 

6 

4 




10 
25 

10 
6 

7 


15 
30 
15 
10 
10 


Brick and Stone. . . 
Cast Iron 


Wrought Iron 

Steel 





1363. Twice as much strain is caused by a suddenly 
applied stress as by one that is gradually applied. For this 
reason a larger safety factor is used for shocks than for 
steady stresses. In general, the factor of safety for a given 
material must be chosen according to the nature of the 
stress. 

The designer usually chooses his own factors of safety. 
If the material has been tested, or the specifications call for 
a certain strength, then the factor of safety can be chosen 
accordingly. 

Example. — Assuming the mortar and brick to be of the same 
strength, how many tons could be safely laid upon a brick column 
2 feet square and 8 feet high ? 

Solution.— P ^ A 6*2 = 2 X 2 x 144 x 2,500 = 1,440,000 lb. = 720 
tons. The factor of safety for this case is 15 (see Art. 1362 and 
Table 28) ; hence, 720 h- 15 = 48 tons. Ans. 

Example, — What must be the diameter of the journals of a wrought 
iron locomotive axle to resist shearing safely, the weight on the axle 
being 40,000 pounds ? 



AS^ 



, or A = 



Solution. — Let /"be the factor of safety; then, jP 

J 

Pf 

-—-. Since the axle has two journals, the stress on each journal is 

20,000 lb. Owing to inequalities in the track, the load is not a steady 
one, but varies ; for this reason, the factor of safety will be taken as 6. 

Then, A = ^^'^^^ >< « 

Ans. 



50,000 



= 2.4 sq. in. Therefore, ^=4/-^ = If". 

r .7854 



758 STRENGTH OF MATERIALS. 

Example. — Considering the piston rod of a steam engine as if its 
length were less than ten times its diameter, what must be the 
diameter of a steel rod, if the piston is 18 inchen in diameter and the 
steam pressure is 110 pounds per square inch ? 

Solution.— Area of piston is 182 x .7854 = 254. 47 sq. in. 254.47 X 110 = 

27,991.7 lb. , or, say, 28,000 lb. = stress in the rod. A =^ = '^'^'?^^^}^ 

Oa 150,000 

/ 1 87 
— 1.87", nearly. Hence, diameter =4/ ^ = 1.543', say l^V* Ans. 

1364. When designing a machine, care should be taken 
(1) to make every part strong enough to resist any stress likely 
to be applied to it; ajid {^) to make all parts of equal strength. 

The reason for the first statement is obvious, and the 
second should be equally clear, since no machine can be 
stronger than its weakest part (proportioned, of course, for 
the stress it is to bear), and those parts of the machine 
which are stronger than others contain an excess of material 
which is wasted. In actual practice, however, this second 
rule is frequently modified. Some machines are intended 
to be massive and rigid, and need an excess of material to 
make them so ; in others, there are difficulties in casting that 
modify the rule, etc. , etc. In most cases, the designer must 
rely on his own judgment. 



EXAMPLES FOR PRACTICE. 

1. A cast iron bar is subjected to a steady tensile stress of 120,000 
pounds. The cross-section is an ellipse whose axes are 6 and 4 inches. 
{a) What is the stress per square inch ? (Jf) What load will the bar 
carry with safety ? A i ^'^^ 6,366.18 lb. per sq. in. 

• 1 {b) 62,832 lb. 

2. How much will a piece of steel 2 inches square and 10 inches 
long shorten under a load of 300,000 pounds ? Ans. .025'. 

3. A cylindrical wooden pin If inches in diameter is subjected to a 
double shearing stress. If the stress is suddenly applied, what total 
force is necessary to shear the pin ? Ans. 7,216 lb. 

4. A wrought iron tie rod is f inch diameter; how long must it 
be to lengthen f inch under a steady pull of 5,000 pounds ? Ans. 69 ft. 

5. A steel bar having a cross-section of 5" X 4" and 14 feet long is 
lengthened .036 inch by a steady pull of 120,000 pounds; what is its 
coefficient of elasticity ? Ans. 28,000,000 lb. per sq. in. 



STRENGTH OF MATERIALS. 



759 



6. Which is the stronger, weight for weight, a bar of chestnut wood 
whose tensile strength is 12,000 pounds per square inch and specific 
gravity .61, or a bar of steel whose tensile strength is 125,000 pounds 
per square inch? 

7. What should be the diameter of a cast-iron pin subjected to a 
suddenly applied double shearing stress of 40,000 pounds to with- 
stand the shocks with safety ? Ans. 4f ", nearly, 

8. What safe steady load may be placed upon a brick column 2 feet 
square and 9 feet high ? Ans. 96,000 lb. 

PIPES AND CYLINDERS. 

1365. A pipe or cylinder subjected to a pressure of 
steam or water is strained equally in all its parts, and, when 
rupture occurs, it is in the direction of its length. 

Let d^ inside diameter of pipe in inches; 
/= length of pipe in inches; 
p = pressure in pounds per square inch; 
Pi= total pressure. 

Then, P = p Id. 

This formula is derived from a principle of hydrostatics 
that the pressure of water in any direction is equal to the 
pressure on a plane perpen- 
dicular to that direction. In 
Fig. 317, suppose the direction 
of pressure to be as shown by 
the arrows; A B would then 
be the plane perpendicular to 
this direction, the width of the 
plane being equal to the diam- 
eter, and the length equal to 
the length of the pipe. The 
•area of the plane would then 
be I X d, and the total pres- 
sure P^^p X I X d, 2iS above. 

Suppose the pipe to have a 
thickness t, and let 5 be the Fig. sir. 

working strength of the material; then, the resistance of 
the pipe on each side is t I S. Resistance must equal 
pressure ; therefore, pld^^^tlS^ or 

pd = %tS. (111.) 




760 STRENGTH OF MATERIALS. 

Also, p = — r-, which is the maximum pressure a pipe of 
^ a 

a given material and of given dimensions can stand. 

The pressure of water per square inch may be found by 

the formula, / =: .434 /i, where /i is the head in feet. In 

pipes where shocks are likely to occur, the factor of safety 

should be high. The thickness of a pipe to resist a given 

pressure varies directly as its diameter, the pressure 

remaining constant. 

Example. — Find the factor of safety for a cast-iron water pipe 
12 inches in diameter and f inch thick, under a head of 350 feet. 

Solution. — Here /, pressure per square inch, equals .434^ = 
.434 X 350 = 151.9 lb. Substituting, in formula 111, the values given, 
151.9 X 12 = 2 X I X 5, or 5= 1,215.2 lb. per sq. in. 

In Table 25, Art. 1356, the ultimate tensile strength of cast iron 
is given as 20,000 pounds per square inch; then, the factor of safety 

is/— ^ J-,- r > — IQ +♦ Ans. The pipe would, therefore, be secure 
•^ 1, 210.2 

against shocks. 

Example. — Find the proper thickness for a wrought-iron steam pipe 
18 inches in diameter to resist a pressure of 140 pounds per square inch. 

Solution. — Using a factor of safety of 10, the working strength S = 

55,000 ..^AiK • -p f 1 1.^ , ;^^ 140X18 

-— — — = 5,500 lb. per sq. m. From formula 111, / = ^^ = -^ — FTTm 
10 /*o /i X o,oUU 

= 0.23 in. In practice, however, the thickness is made somewhat 

greater than the formula requires. 



CYLINDERS. 

1366. The tendency of a cylinder subjected to internal 
pressure is to fail or rupture in the direction of its length, 
the same as a pipe. 

Let TT (pronounced//) be the ratio of the circumference 
of a circle to its diameter = 3. 1416 ; then, ^ r = . 7854 = ratio 
of area to the square of the diameter. 

The total pressure on the cylinder head = |- tt t^' /. Let 

S = working unit stress = -^, then tt d ^ S = the resistance 
to rupture caused by the pressure acting on the opposite 



STRENGTH OF MATERIALS. 



761 



cylinder heads and tending to elongate the cylinder. Since 
the resistance must equal the force, or pressure, \t, d"^ p ■=^ 



t: d t S^ or 



Also, / = 



(112.) 



4/5 
d ' 



2fS 



Since, for longitudinal rupture, / = — -j—, it Is seen that 

a cylinder is twice as strong against transverse rupture as 
against longitudinal rupture. Hence, other things being 
equal, the cylinder will always fail by longitudinal rupture. 

1367. The foregoing formulas are for comparatively 
thin pipes and cylinders, in which the thickness is less than 
about -^Q inside radius. For pipes and cylinders whose 
thickness is greater than ^^^ radius, use the following 
formula, in which r = the inner radius, and the other 
letters have the same meaning as before. 



/ = 



r-Yf 



(113.) 



Substituting the values given in the example in Art. 1 365, 
in formula 113, instead of formula 111, 



/ 



St 



or vS 



/(r + /)_ 151.9 X (6 + f) 



151.9 X Gf X 1= 1,307.1 lb. 

When formula 111 was vised, 5=1,215.2 lb.; hence, 
formula 113 gives, for this case, a value Vl\io^ or \ greater. 

The formula for spheres i3 the same as that for transverse 
rupture of cylinders, or p d=^ 4, t S. 

1 368. A cylinder under external pressure is theoreti- 
cally in a similar condition to one under internal pressure, 
so long as its cross-section remains a true circle. A uniform 
internal pressure tends to preserve the true circular form, but 
an external pressure tends to increase the slightest variation 
from the circle, and to render the cross-section elliptical. 
The distortion, when once begun, increases rapidly, and 



762 STRENGTH OF MATERIALS. 

failure occurs by the collapsing of the tube rather than by 
the crushing of the material. The flues of a steam boiler 
are the most common instances of cylinders subjected to 
external pressure. 

The letters having the same meaning as before, the follow- 
ing formula gives the collapsing pressure in pounds per 
square inch for wrought-iron pipe: 

/= 9,600,000^. (114.) 

Example. — What must be the thickness of a boiler tube 2 inches in 
diameter and 11 feet long, if the steam pressure is to be not over 160 
pounds per square inch ? 

Solution. — Using formula 114, with a factor of safety of 10, and 
solving for /, 



/= y/ ^^pld _^-|y iO X 160 X 11 X 12 X2 _°-|7ir 
r 9,600,000 ~ r 9,600,000 ~ r 250* 

Hence, log t = ^"^S '^'^ -^og 250 ^y gr^^^^g^ ^^ ^ ^ .2386", say i". 

<*. lo 



EXAMPLES FOR PRACTICE. 

1. What must be the thickness of a 16-inch cast-iron stand pipe 
which is subjected to a head of water of 250 feet ? Assume that the 
stress is steady. Ans. .26". 

2. What should be the thickness of a wrought-iron boiler flue 15 
feet long, 4 inches in diameter, and subjected to an external pressure 
of 200 pounds per square inch ? Ans. .42". 

3. What pressure per square inch can be safely sustained by a cast- 
iron cylinder 12 inches in diameter and 3 inches thick ? 

Ans. 1,111^ lb. per sq. in. 

4. What external pressure per square inch can a wrought iron pipe 
20 feet long, 3 inches in diameter, and -f inch thick, safely sustain and 
be secure against shocks ? Ans. 157.2 lb. per sq. in. 

5. A cast-iron cylinder 14 inches in diameter sustains a total pres- 
sure of 125 tons ; what is the necessary thickness, assuming that the 
pressure is gradually applied, and that the cylinder is not subjected to 
shocks? Ans. 6.65". 

6. A cylindrical boiler shell 3 feet in diameter is subjected to a 
steady hydrostatic pressure of 180 pounds per square inch. What 
should its thickness be if made of steel having a tensile strength of 
60,000 pounds per square inch ? Ans. .27". 



STRENGTH OF MATERIALS. 



763 



ELEMENTARY GRAPHICAL STATICS. 

Before taking up the subject of flexure, some fundamen- 
tal principles of Graphical Statics not heretofore considered 
will be explained and applied to the case of beams. 



FORCE DIAGRAM AND EQUILIBRIUM POLYGON. 

1369. In Arts. 878 and 879, the polygon of forces 
was used to find the resultant of several forces having a 
common point of application, or whose lines of action 
passed through a common point. A method of finding the 
resultant will now be given when the forces lie, or may be 
considered as lying, in the same plane, but their lines of 
action do not pass through a common point. 

In Fig. 318, let F^, F^^ and F^ be three forces whose mag- 
nitudes are represented by the lengths of their respective 
lines, and their directions by the positions of the lines and by 




Scale 1=80 Ib^ 



Fig. 318. 

Construct the polygon of forces O 1 2 S O 
in the same manner as described in 
Art. 878, O 3 representing the direction and magnitude of 
the resultant. Everything is now known except the line on 



the arrow-heads. 

as shown at [d) 



764 STRENGTH OF MATERIALS. 

which the point of application of the resultant must lie. To 
find this, proceed as follows: 

Choose any point P, and draw P O^ P 1, P 2^ and P 3. 
Choose any point b^ on the line of direction of one of the 
forces as F^^ and draw lines through b parallel to P O and 
P i, the latter intersecting F^, or F^ prolonged, in c. Draw 
^<^ parallel to P 2^ and intersecting Z^^, oi" -^a prolonged, in d. 
Draw d e parallel to P 8, intersecting the line a b e^ parallel 
to P O, in e. The point ^ is a point on the line of direction 
of the resultant of the three forces. Hence, through ^, 
draw R parallel and equal to O 3 and acting in the same 
direction ; it will be the resultant. 

The method just described is applicable to any number of 
forces considered as acting in the same plane. The result- 
ant can also be found when the forces act in different 
planes, but the method of finding it will not be described 
here. 

The point P is called the pole ; the lines P O, P 1, P 2, 
P 3 joining the pole with the vertexes of the force polygon 
are called the strings or rays ; the force diagram is 
the figure composed of the force polygon, 12 3 O, the 
pole, and the strings. The polygon b e d c b is called the 
equilibrium polygon. 

Since the pole P may be taken anywhere, any number of 
force diagrams and equilibrium polygons may be drawn, all 
of which will give the same value for the resultant, and 
whose lines d e and a e will intersect on the resultant R. 
To test the accuracy of the work, take a new pole and pro- 
ceed as before. If the work has been done correctly, d e 
and a e will intersect on R. 

The equilibrium polygon gives an easy method of resolving 
a force into two components. 

Example. — In Fig. 319, let F= 16 pounds be the force, and let it be 
required to resolve it into two parallel components, A and B^ at dis- 
tances respectively of 5 feet and 15 feet from F. What will be the 
magnitudes of A and B ? 

Solution. — Draw O Ito represent F^= 16 lb. Choose any conven- 
ient pole 1\ and draw the rays P O and P 1. Take any point a on F 



STRENGTH OF MATERIALS. 



7G5 



and draw ab parallel to PO, intersecting A in b, and ac parallel to 
P 1, intersecting B in c. Join b and c by the line b c. Through the 
pole P, draw P d parallel to b c, intersecting O 1 in d. Then O d is the 
magnitude of A, measure, to the scale to which 6>i was drawn, and 
^i is the magnitude of B to the same scale. 




Fig. 319. 

ISTO. If the components are not parallel to the given 
force, they must intersect its line of direction in a common 
point. 

In Fig. 320, let /^= 16 pounds be the force; it is required 
to resolve it into two compo- 
nents A and B, intersecting at a, 
as shown. Draw O 1 to some 
convenient scale equal to 16 
pounds; then draw O Pand 1 P 
parallel to A and B^ and O P 
and P 1 are the values of the 
components A and B^ respec- 
tively, both in magnitude and direction. 





Fig. 320. 



Example. — Let Fx, Fi, Fz, 7^4, and F-,, Fig. 321, be five forces whose 
magnitudes are 7, 10, 5, 12, and 15 pounds, respectively. It is required 
to find their resultant and to resolve this resultant into two com- 
ponents parallel to it and passing through the points a and b. 

Solution. — Choose any point O, Fig. 321, and draw 01 parallel and 
equal to Fx ; 1-2 parallel and equal to F^, etc. ; O 5 will be the value of 
the resultant, and its direction will be from O to 5, opposed to the 
other forces acting around the polygon. Choose a pole Py and com- 
plete the force diagram. Choose a point ^ on Fi, and draw the equilib- 
rium polygon cdefghc, the intersection oi c k, parallel to PO, and 
g hy parallel to P5, gives a point k, on the resultant R. Through h, 
draw R parallel to 5, and it will be the position of the line of action 



766 



STRENGTH OF MATERIALS. 



of the resultant of the five forces. The components must pass through 
the points a and b, according to the conditions ; hence, through a and 
b, draw Vx and F2 parallel to R. Since O 5 represents the magnitude 
cf i?, draw kk and /^/ parallel to P O and P5, respectively, as in Fig. 




Fig. 321. 

319 (they, of course, coincide with c h and g k, since the same pole P is 
used), intersecting Vi and V-i in k and /. Join k and /, and draw P Q 
parallel to k I. Then, O Q= V„ and Q5= V^. 



COMPOSITION OF MOMENTS. 
1371. In Art. 906 it was stated that the moment of a 
force about a point is the product of the magnitude of the force 
by the perpendicular di'stance from the point to the line of 
action of the force. A force can act in two ways upon a 
body : it can either produce a motion of translation — that is, 
cause all the points of the body to move in straight parallel 
lines — or it can produce a motion of rotation — that is, make 
the body turn. A moment measures the capacity of a force 
to produce rotation about a given point. For example. 




STRENGTH OF MATERIALS. 767 

suppose, in Fig! 322, that A C is a lever 30 inches long, 

having a fulcrum at B 10 inches from A. If a weight is 

suspended from C, it will cause the bar to rotate about B in 

the direction of the arrow. A weight suspended from A will 

cause it to revolve in the opposite direction, as indicated by 

the arrow. Suppose, for simplicity, that the bar itself weighs 

nothing. If two weights of 12 pounds each are hung 

at A and C^ it is evident that the bar will revolve in the 

direction of the arrow at C, on account of the arm B C being 

longer than the arm A B, Let the weight at A be increased 

until it equals 24 pounds. The bar will then balance exactly, 

and any additional \^ iq!L—^ ^0— 

weight at A will cause I ^ | ^ 

the bar to rotate in the * ^ » 

opposite direction, as /^^ 

shown by the arrow at ^-^ 

.1 , • , -rxTi ., 24lh8. 12 lbs, 

that pomt. When the fig. 322. 

lever is balanced, it will be found that 24 X 10 = 12 X 20, 
or, considering B as the center of moments, 24 X per- 
pendicular distance A B = 12 X perpendicular distance 
B C. In other words, the moment of W about B must 
equal the moment of P about B — that is, the two moments 
must be equal. Further, intends to cause rotation in the 
direction in which the hands of a watch move, and will, for 
convenience, be considered positive, or -|- ; I^K tends to cause 
rotation in a direction opposite to the hands of a watch, and 
will be considered negative, or — . Adding the two algebrai- 
cally, Px^ C-\-{~ WxA B) = PxB C- WxA B = Q), 
since the two moments are equal. Hence the following general 

Rule. — 0?ie of the necessary conditions of equilibrium is 
that the algebraic sunt of the ^moments of all the forces about 
a given point should equal zero. 

Applications of this rule will occur farther on. 



GRAPHICAL EXPRESSIONS FOR MOMENTS. 

1372. The moment of a single force may be expressed 
graphically in the following manner : Let i^= 10 pounds 
be the given force (see Fig. 323), and r, at a distance from 



708 



STRENGTH OF MATERIALS. 



F=:fc=z 7-2 feet, be the center of rotation (center of mo- 
ments). Draw O 1 parallel to F and equal to 10 pounds. 
Choose any point P as a pole, and draw the rays P O and 
P 1 ; also draw P2 perpendicular to O 1. Through any point 
b on F, draw the sides a b and g b of the equilibrium poly- 
gon ; they correspond to <^ ^ and d e^ Fig. 318, through the 
intersection of which the resultant must pass, the resultant 
F being given in the present case. Prolong a b^ and draw 




^>p 



Fig. 323. 

e </ through c parallel to /% intersecting b g and a b vn d and 
e. It can now be shown that the moment of F about c = 
P2 X d e^ when P 2 is measured to the same scale as O i, 
and e d'ls measured to the same scale as c f. 

The line P 2 is> called the pole distance and will here- 
after be always denoted by the letter H, The line d e is 
called the intercept. Hence, the pole distance multiplied 
by the intercept equals the moment, or, denoting the inter- 
cept by jK, moment = H y. 

The statement made in the last sentence is one of the 
most important facts in Graphical Statics, and should be 
thoroughly understood. In the triangle POl, the lines 
PO and Pi represent the components of the force F in the 
directions a b and g b^ respectively, while the lines of action 
of those components are a b and g b, meeting at b. As d e 
is limited hy g b and a b (produced), we may give the fol- 
lowing definition : The intercept of a force whose moment 
about a point is to be found is the segment (or portion) 
which the two components (produced, if necessary) cut ofl 



STRENGTH OF MATERIALS. 769 

from a line drawn through the center of moments parallel to 
the direction of the force. 

1373. The pole distance and intercept for the moment 
of several forces about a given point may be determined in 
a similar manner, by first finding the magnitude and position 
of the resultant of all the forces; the moment of this result- 
ant about the given point will equal the value of the 
resultant moment of all the forces which tend to produce 
rotation about that point. 

Example. — Let Fi = 20 pounds, F2 = 25 pounds, and Fa = 18 pounds 



^P 



Fig. 324. 

be three forces acting as shown in Fig, 324; find their resultant moment 
about the point C. 

Solution. — Draw the force diagram, equilibrium polygon, and re- 
sultant i? as previously described. Draw dC^ parallel to 7?. The 
i itercept d e, multiplied by the pole distance PQ = th.e resultant 
moment. 

137'4. If all the forces are parallel, the force polygon 
is a straight line. This is evident, since if a line of the 
force polygon be drawn parallel to one of the forces, and 
from one end of this line a second line be drawn parallel 
to another force, the second line will coincide with the 
first. 

Example, — Let Fi = 30, 7^2 = 20, and F3 = 20, all in pounds, be three 
parallel forces acting downwards,, as shown in Fig. 325. It is required 
to find their resultant moment (algebraic sum of the moments) and the 
moment of their resultant, all moments to be taken about the point C 

D. 0. III.— 7 



770 



STRENGTH OF MATERIALS. 



Solution.— Lay off Ol = Z0 lb. = Fi, 1-2 — 20 lb. = T^a and 2-S — 20 
lb. = Fz, and O S vs, the value of the resultant. Choose some point P as 
a pole and draw the rays. Take any point, as b, on any force, as Fx, 
and complete the equilibrium polygon be deb; then, the line of action 
of the resultant must pass through e. Through C draw C i parallel to 
R and prolong de. The moment of R about C equals the pole distance 
//, multiplied by the intercept h /, since // / is that part of the line drawn 
through C parallel to R, and included between the lines be and d e oi 
the equilibrium polygon which meet upon R. Measuring hi to the 
scale of distances (1 in. = 40 ft.), it equals 23 ft. Measuring H to the 
scale of forces, it equals 40 lb. Consequently, the moment of R about 




Scale of forces 1=40 lb. 
Scale of distance 1=40^ 

Fig. 325. 



JE 



C= 40 X 23 = 920 ft. -lb. Considering the force Fa, the intercept is ^/, 
since F3 is parallel to R, and, consequently, to Cz; also, ^/is that part 
of the line C/ included between the sides ed a.nd de, which meet on F3. 
Measuring £■ i, it is found to equal 28 ft. Hence, the moment of F3 
about C = 40 X 28 = 1,120 ft. -lb. The moment of iS about C = ^/X/^ 
= 40 X 13 = 520 ft. -lb. The moment of Fx = Hx/ h = 4:0 X IS = 720 
ft. -lb. Now 7^3 and F2 have positive moments, since they tend to cause 
rotation in the direction of the hands of a watch, while Fi has a nega- 
tive moment, since it tends to cause rotation in the opposite direction. 
Consequently, adding the moments algebraically, the resultant moment 
= 1,120 + 520 — 720 = 920 ft.-lb.^ the same as the moment of the 
resultant. 



STRENGTH OF MATERIALS. 771 

Having described the fundamental principles of Graphical 
Statics, the subject of Strength of Materials will now be 
continued, and the stresses due to flexure and torsion 
discussed. 

BEAMS. 

1375. Any bar resting upon supports in a horizontal 
position is called a beam. 

1376. A simple beam is a beam resting upon two 
supports very near its ends. 

1377. A cantilever is a beam resting upon one sup- 
port in its middle, or which has one end fixed (as in a wall) 
and the other end free. 

1378- A restrained beam is one which has both 
ends fixed (as a plate riveted to its supports at both ends). 

1 379. A continuous beam is one which rests upon 
more than two supports. 

In this Course, the continuous beam will not be discussed, 
as the subject requires a knowledge oi higher mathematics. 



REACTIONS OF SUPPORTS. 

1380. All forces which act upon beams will be con- 
sidered as vertical, unless distinctly stated otherwise. Ac- 
cording to the third law of motion, every action has an 
equal and opposite reaction. Hence, when a beam is acted 
upon by downward forces, the supports react upwards. It 
is required to find the value of the reaction at each support. 
If a simple beam is uniformly loaded or has a load in the 
middle, it is evident that the reaction of each support is 
one-half the load, plus one-half the weight of the beam. If 
the load is not uniformly distributed over the beam, or if 
the load or loads are not in the middle, the reactions of the 
two supports will be different. The upward reactions are 
considered positive, and the downward forces negative. Ic 



772 



STRENGTH OF MATERIALS. 



order that a beam may be in equilibrium, three conditions 
must be fulfilled: 

I. The algebraic sum of all vertical forces = 0. 

II. The algebraic sum of all horizontal forces = 0. 

III. The algebraic sum of the moments of all forces 
about any point = 0. 

Since the loads act downwards and the reactions upwards, 
the first condition states that the sum of all the loads must 
equal the sum of the reactions of the supports. 

Example. — Let 7?i be the reaction of the left support, R-z the reac- 
tion of the right support, and the distance between the two supports 
14 feet. Suppose that loads of 50, 80, 100, 70, and 30 pounds are placed 




Scale 1-^160 lb- 



Fig. 326. 

on the beam at distances from the left support equal to 2, 5, 8. 10, and 
12| feet, respectively. Required the reactions of the supports, neg- 
lecting the weight of the beam. See Fig. 326. 



ti' 



STRENGTH OF MATERIALS. 773 

Solution. — The reactions may be found graphically by resolving 
the resultant of the weights (which, in this case, acts vertically down- 
wards) into two parallel components, passing through the points 
of support. The sum of the reactions is equal to the sum of the com- 
ponents, but the two sums have different signs. Draw the beam to 
some convenient scale, and locate the loads as shown in the figure. 
Draw the force diagram, making O i = 50 lb., i-^ = 80 lb., etc., O 5 
representing the force polygon. 

Choose a point b on the line of action of the force Fi, and draw the 
equilibrium polygon abcdefga;ab and/ ^ intersect at //, the 
point through which the resultant R must pass. Draw P x parallel to 
a g, and O .r will be the reaction (= component) 7?i, and x 5 the reac- 
tion Ri. Measuring O jrand x 5 to the same scale used to draw O 5, 
we find 7?i = 161 lb., and ^^ = 169 lb. By calculation, 7?i = 160.4 lb., 
and y?2 = 169.6 lb. This shows that the graphical method is accurate 
enough for all practical purposes. The larger the scale used, the more 
accurate will be the results. 

The reactions and forces acting upwards will always be 
considered as positive, or -j-, ^i^d the downward weights as 
negative, or — . It is plain that the first condition of 
equilibrium is satisfied when the sum of the positive forces 
and reactions is equal to the sum of the negative forces. 



THE VERTICAL SHEAR. 

1381. In Fig. 326, imagine that part of the beam at a 
minute distance to the left of a vertical line passing through 
the point of support A, to be acted upon by the reaction R^ = 
160 pounds, and that part to the right of the line to be acted 
upon by the equal downward force due to the load. The two 
forces acting in opposite directions tend to shear the beam. 

Suppose the line had been situated at the point marked 3 
instead oi a.t A; the reaction upwards would then be partly 
counterbalanced by the 50-pound weight, and the total 
reaction at this point would be 160 — 50 = 110 pounds. 
Since the upward reaction must equal the downward load at 
the same point, the downward force at 3 also equals 110 
pounds, and the shear at this point is 110 pounds. At the 
point 6, or any point between 5 and <5', the downward force 
due to the weight at the left is 50 -f- 80 = 130 pounds, and 
the upward reaction is 160 pounds. The resultant shear is 



774 



STRENGTH OF MATERIALS. 



therefore, 160 — 130 = 30 pounds. At any point between 
8 and 10, the shear is IGO — (50 + 80 -(- 100) = — 70 pounds. 
The negative sign means nothing more than that the weights 
exceed the reaction of the left-hand support. 

T/ie vertical shear equals the reaction of the left-hand 
support, minus all the loads on the beam to the left of the pomt 
considered. 

For a simple beam, the greatest positive shear is at the 
left-hand support, and the greatest negative shear is at the 
right-hand support, and both shears are equal to the reactions 
at those points. 

1382. The vertical shear may be represented graph!- 




Fig. 327. 



cally, as shown in Fig. 327, which is Fig. 326 repeated. 
Draw the force diagram, continue the lines of action of R^ and 



STRENGTH OF MATERIALS. 775 

R^ downwards and make O' 5' = O 5. Through x draw the 
horizontal line x x\ called the shear axis. The vertical 
shear is the same for any point between A and 2 and =: O x 
= O' x" = 160; hence, draw O' Ji parallel to x x\ and any 
ordinate measured from x x' to O' hy between O' and h = 
160 pounds = the vertical shear at any point between O' and 
h when O' h ^= A 2. Through 1, draw 1 k and project the 
points 2 and 5 upon it, in i and k. Then, the length of the 
ordinate between x x' and i k^= the vertical shear between 
2 and 5. In the same way, find the remaining points /, m, etc. 

The broken line O' h i .../ is called the shear line; 

and the figure O' h i r s t x' x" is called the shear 

diagram. To find the shear at any point, as 11, project 
the point upon the shear axis and measure the ordinate to 
the shear line, drawn through the projected point. If the 
ordinate is measured from the shear axis upwards, the ver- 
tical shear is positive ; if downwards, it is negative. For the 
point 11, the vertical shear = — 140 lb. The maximum 
negative vertical shear is — 170 Vo. ^^ x' t =^ x 5. The 
greatest shear, whether positive or negative, is the one 
which the beam must be designed to withstand. 

1383. A beam seldom fails through shearing, but 
generally breaks by reason of the load bending and breaking 
it; that is, through flexure. In order to design a beam to 
resist flexure, the greatest (maximum) bending moment 
must be known. 

The bending moment at any point of a beam is the alge- 
braic Sinn of the moments of all of the forces {the reaction 
included) acthig upon the beam, on either side of that pointy 
the point being considered as the center of moments. 

The expression "algebraic sum" refers to the fact that, 
when considering the forces acting at the left of the point 
taken, the moments of all the forces acting upwards are 
considered positive, and the moments of all the forces acting 
downwards are considered negative. Hence, the algebraic 
sum is the moment of the left reaction about the given point, 
minus the sum of the moments about the same point of all 



776 STRENGTH OF MATERIALS. 

the downward forces between the reaction and the given 
point. Should there be any force or forces acting upwards, 
their moments must be added^ since they are positive. If 
the forces on the right of the point are considered, all lever 
arms are negative, distances to the left of the point being -|-, 
to the right, — (see Art. 1371). Hence, the downward 
forces give positive moments, and the right reaction gives a 
negative moment. This is as it should be, for the downward 
forces on the right and the upward forces on the left tend 
to rotate the beam in the direction of the hands of a watch, 
while the downward forces on the left and the upward forces 
on the right tend to rotate the beain in the opposite direction. 

1384. To find the bending moment for any point of a 

beam, as 7 in Fig. 327, by the graphical method, draw the 

vertical line 1-T through the point. Let y = that part of 

the line included between a g and a c f g of the equilibrium 

polygon ( = vertical intercept). Then, i7 x JK = the bending 

moment. //, of course, = the pole distance == P ti. For 

any other point on the beam, the bending moment is found 

in the same manner — i. e., by drawing a vertical line through 

the point and measuring that part of it included between 

the upper and lower lines of the equilibrium polygon. The 

scale to which the intercept y is measured is the same as that 

used in drawing the length A B oi the beam. The pole 

distance H is measured to the same scale as O 5. In the 

present case, y = 2.05 feet, and i/= 349 pounds; hence, the 

bending moment for the point 7 is Hy = 349 X 2.05 = 715.45 

foot-pounds. 

Note. — The expression "foot-pounds," used in stating the value of 
a moment, must not be confounded with foot-pounds of work. The 
former means simply that a force has been multiplied by a distance, 
while the latter means that a resistance has been overcome through a 
distance. In expressing the value of a moment, the force is usually- 
measured in pounds or tons, and the distance in inches or feet ; hence, 
the moment may be inch-pounds or inch-tons and foot-pounds or foot- 
tons. Unless otherwise stated, the bending moment will always be 
expressed in inch-pounds, the length of the beam being always 
measured in inches, and, consequently, also the length of the inter- 
cept y. 

1385. If expressed in inch-pounds, the value of the 
moment just found is 715.45 X 12 = 8,585.4 inch-pounds. 



STRENGTH OF MATERIALS. 777 

It will be noticed that after the force diagram and equilib- 
rium polygon have been drawn the value of the bending 
moment depends solely upon the value of j, since the lengtti 
P 11 := H is fixed. At the points a and g^ directly under the 
points of support of the beam, jj/ = ; hence, for these two 
points, bending moment = H y = H X = 0; that is, for any 
simple beam, the bending moment at either support is zero. 
The greatest value for the bending moment will evidently be 
at the point <9, since d 8' is the longest vertical line which 
can be included between a g and a c f g. 

The figure a c e g a is called the diagram of bending 
moments. 

1 386. Consider now the case of a simple beam uniformly 
loaded. Let the distance between the supports in Fig. 328 
be 12 feet, and let the total load uniformly distributed over 
the beam be 21G pounds. Divide the load into a convenient 
number of equal parts, the more the better, say 12, in this 
case. The load which each part represents is 216 -^ 12 = 18 
pounds. For convenience, lay ofif O C on the vertical 
through the left-hand support, equal to 21G pounds to the 
scale chosen, and divide it into 12 equal parts, O a, a b, etc. ; 
each part will represent 18 pounds to the same scale. 
Choose a pole P, and draw the rays P O^ P a^ P b, etc. 
Through the points d, e^ etc., the centers of gravity of the 
equal subdivisions of the load, draw the verticals d i, e <?, 
/ 5, fctc, intersecting the horizontals through (9, a^ b^ etc., 
in i, 3, 5, etc. Draw O 1, 1-2^ 2-3, 3-J,., Jf-5, etc. , and the 
brokcxi line thus found will be the shear line. In drawing 
the shear line for a uniform load in this manner, it is assumed 
that each part of the total load is concentrated at its center 
of gravity, or, in other words, that a force equal to each 
small load (18 lb.) acts upon the beam at each of the points 
d, e, /", etc. 

Construct the diagram of bending moments in the ordi- 
nary manner by drawing g i parallel to P O, i k parallel to 
Pa, etc. Draw P M parallel to g h, and M q horizontal; 
M Q IS the shear axis. When the load is uniform and the 



778 



STRENGTH OF MATERIALS. 



work has been done correctly, O M should equal M 6"— that 
is, the reactions of the two supports are equal. 

13S7. The shear line is not a broken line in reality, as 
shown, since the load is distributed evenly over the entire 
beam, and not divided into small loads concentrated at </, e^ /", 
etc., as was assumed. The points i, S^ etc., are evidently 




''" Scale, of forces X^t^O 11^ 
Scale of distance 1=4^. 

Fig. 328. 

too high, and the points^, ^ etc., too low. To find the 
real shear line, bisect the lines 1-2, S-Jf^ etc., locating the 
points 6, 7, 8, etc. Trace a line through (9, 6, 7, 

8 A^, and it will be the real shear line. For all cases of 

a uniform load, the shear line will be straight and may be 
drawn from O to N directly. 

The diagram of bending moments is also not quite exact, 
but may be corrected by tracing a curve through the points 
g and /?, which shall be tangent to g i, i k, k /, etc., at their 
middle points, as shown in the figure. 



STRENGTH OF MATERIALS. 779 

1388. To find the maximum bending moment for any 
beam having two supports, draw a vertical line through r, 
where the shear line cuts the shear axis, and the intercept, 
S t = y, on the diagram of bending moments, will be the 
greatest value of j, and, consequently, the greatest bending 
moment =z H X y. In the present example, ^ = P ic— 247. 5 
pounds and maximum jj/ = ^^ = 15.72 inches; hence, the 
maximum bending moment — H y — 247. 5 X 15. 72 = 3,890. 7 
inch-pounds. The above is true, no matter how the beam 
may be loaded. In Fig. 327, the shear line cuts the shear 
axis at ^, and d 8\ on the vertical through z. was shown 
previously to be the maximum intercept. 

1389. If there is a uniform load on the beam and one 
or more concentrated loads, as in Fig. 329, the method of 
finding the moment diagram and shear line is similar to that 
used in the last case. In Fig. 329, let the length A B oi the 
beam be 15 feet; the uniformly distributed load 180 pounds, 
with two concentrated loads, one of 24 pounds, 5 feet from 
A^ and the other of 30 pounds, 11 feet from A. Draw the 
beam and loads as shown, the length of the beam and the 
distances of the weights from A being drawn to scale. Divide 
the uniform load into a convenient number of equal parts, 
say 10 in this case ; each part will then represent ^-^- = 18 
pounds. Draw ^ (9 (T, as usual, and lay off 3 of the 18-pound 
subdivisions from O downwards; then lay off 24 pounds, to 
represent the first weight. Lay off four more of the equal 
subdivisions, and then the 30-pound weight. Finally, lay 
off the remaining three equal subdivisions, the point C being 
the end of the last 18-pound subdivision. O C should then 
equal 180 -j- 24 -j- 30 = 234 pounds to the scale to which the 
weights were laid off. It will be noticed in the above that 
the equal subdivisions of the load were laid off on O C until 
that one was reached on which the concentrated loads rested, 
and that the concentrated loads were laid off before the 
equal subdivision on which the concentrated load rested was 
laid off. Had one of the concentrated loads been to the 
right of the center of gravity of the subdivision on which it 



780 



STRENGTH OF MATERIALS. 



rests, the weight of the subdivision would have been laid off 
first. Locate the centers of gravity of the equal subdivisions 
and draw the verticals through them as in the previous case. 
Choose a pole P, draw the rays, the diagram of bending 
moments, and the shear axis M q^ as previously described. 




^L'^ Fig. 329. 

To find the maximum bending moment, the shear line 
must be drawn and its intersection with the shear axis 
determined. 

The weight of the uniform load per foot of length is 
X^ = 12 lb. The weight of that part between A and the 
center of the 24-lb. weight is 12 X 5 = GO lb. Lay off 
(9 i = GO lb. and draw 1-2 horizontal, cutting the vertical 
through the center of the 24-lb. weight in 2. Draw O 2 and 
it will be a part of the shear line. Lay off 2-S vertically 
downwards, equal to 24 lb., locating the point 3. Lay off 
O Jf equal to 12 X 11 + 24 = 15G lb., and draw the horizontal 
^-5, cutting the vertical through the center of the 30-lb. 



STRENGTH OF MATERIALS. 781 

weight in 5. Join S and 5 by the straight line 3-5. Lay off 
5-6 vertically downwards equal to 30 lb. Draw the horizon- 
tal C N, intersecting the vertical B q N in N and join 6 and 
N by the straight line 6 N. The broken line O 2-3-5-6 N is 
the shear line and cuts the shear axis in the point r. Draw- 
ing the vertical r b a^ through r, it intersects the moment 
diagram in a and b\ hence, a b i?> the maximum y. For this 
case, the maximum bending moment =:H x }' =300 X 18.36 
= 5,508 in. -lb. 

It is better, in ordinary practice, to choose the pole Pon a 
line perpendicular to O C and at some distance from O C 
easily measured with the scale used to lay off O C. Thus, 
su.ppose O C to be laid off to a scale of \" = GO lb. At some 
convenient point, as M^ on O C, draw a perpendicular line 
and choose a point on this line whose distance from O C 
shall be easily measurable, say 3^. Then, H is known to 
be exactly 60 X 3^ = 210 lb. and will not have to be meas- 
ured when finding the bending moment H y. 

1390. If the student has familiarized himself with the 
method of constructing the shear and moment diagrams for 
concentrated loads, he will find no difficulty in understanding 
the preceding operations, which may be condensed into the 
following 

Rule. — Divide the beam into an even number of parts (the 
greater the better), and the ttniforni load into Jialf as many. 
Consider the divisions of the load as concentrated loads 
applied^ alternately^ at the various points of division of the 
beam (the ends included); that is^ the first point of division 
{the support) carries no load, the next one does, the following 
one does not, etc. TJien proceed as in the ordinary case of 
concentrated loads. 

Example. — Find the reactions of the supports, the maximum bend- 
ing moments, and the maximum vertical shear of the beam shown in 
Fig. 330, which has one overhanging end. 

Solution. — Draw O C 2sv& the force diagram in the usual manner. 
Construct the bottom curve of the moment diagram in the same man- 
ner as in the preceding cases. The side de is parallel to /'^; e h\^ 
parallel to P C, and cuts the vertical through the right reaction in h. 



782 



STRENGTH OF MATERIALS. 



Join k and g by the straight line g k, and draw P M parallel to g k. 
Then, O Af = 77 lb. = left reaction and MC=11^ lb. = right reaction. 
The shear line is drawn as in the previous cases until the point n, on 
the vertical h n, is reached ; k n here denotes the vertical shear for 
any point between the 50-lb. weight and the right support, and this 
shear is negative. The point k denotes the intersection of the shear 
axis and the vertical through the right support. For any point to the 




Scale, of fortes if^lOO Ih* 
Scale of distance 1^8^ 



Fig. 330. 



right of ^, between ^ and /, the vertical shear is positive, and is equal 
to 70 lb. ; hence, lay o^ k q upwards equal to 70 lb. , and draw q r 
horizontal. The line, O 6-6-7 — ftqr'is the shear line. Measur- 
ing O M,kn, and k q, it is found that O M =11 \b., k n = — 103 lb., and 
^ ^ = 70 lb. ; therefore, k n = ~ 103 lb. = maximum vertical shear ; c i 
i§ evidently the rnaximum y ; hence, the ma^cimum bending moment = 



STRENGTH OF MATERIALS. 



783 



Hy = /* ^ X ^ 2 = 200 X 26" = 5, 200 in. -lb. Any value of y measured in 
the -^oXY^on g a b c d s g is positive, and any value measured in the 
triangle e hs !■=> negative. Consequently, the bending moment for any 
point between s and the vertical, through the center of the 70-lb. 
weight, is negative, since Hx{~y)= — Hy. In all cases, when design- 

56' 




Ji'lG. 331. 



ing beams with overhanging studs, the maximum bending moment, 
whether positive or negative, should be used. In the present case, the 
maximum negative/, or hj, is less than the maximum positive j, or 
ci\ therefore, the maximum negative bending moment is also less 
than the maximum positive moment. 



784 STRENGTH OF MATERIALS. 

Example. — Fig. 331 shows a beam overhanging both supports, 
which carries a uniform load of 15 pounds per foot of length, and has 
five concentrated loads at distances from the supports as marked in 
the figure. Required the reactions of the supports, the maximum 
positive and negative bending moments, and the maximum vertical 
shear. 

Solution. — Construct the force diagram and equilibrium polygon 
in the ordinary manner, continuing the latter to b and a, points on the 
verticals passing through the ends of the beam. Draw (^ /z and ag 
parallel to P C and F O respectively, intersecting the verticals through 
the points of support in A and g. Join g and /i, and draw P B parallel 
to g/t. Then, (9 ^ = 588 lb. = 7?i, and BC= 612 lb. = R^. Through 
B, draw the shear axis 7n q. To draw the shear line, proceed as fol- 
lows: The shear for any point to the left of the left support is 
negative, and for any point to the right of the right support is 
positive ; between the two supports it is positive or negative, accord- 
ing to the manner of loading, and the point considered. The negative 
shear at the left support = 15 X 6 + 100 = 190 lb. ; hence, lay o'S. B d 
downwards equal to 190 lb. For a point to a minute distance to the 
right of e, the shear is 15 X 3 + 100 = 145 lb. = ef, and for a minute 
distance to the left, it is 15 X 3 = 45 lb. ■= e i \ at ?;z, it is 0. Conse- 
quently, 7nzfdis the shear line between the end of the beam and the 
left support. Lay off Ol = Bd=190 lb. and draw the shear line 

l-^-o-4- — n in the usual manner. Draw the shear line 5-6-7 q, 

laying off /&5 = 15 X 8 + 100 = 220 lb. ; /(? = 100 + 15 X 2^ = 137^ lb., and 
6-7 = 100 lb. At q, the vertical shear is again 0, The broken line 

7/1 if d 1-2-3 71 5-6-7 q is the shear line. The maximum positive 

bending moment is HXy = PuX sc = 1,000 X 22.5 = 22,500 in. -lb. 
The greatest maximum negative moment is, Hy,{— y) ■= P uY, — t h-=. 
1,000 X — 11.6 = — 11,600 in. -lb. It will be noticed that there are two 
negative and one positive maximum bending moments. 



1391. The student should now be able to find the 
bending moment for any beam having but two supports, 
whatever the character of the loading. The bending mo- 
ment plays a very important part in the flexure of beams, 
which is the next subject to be considered. In all cases of 
loading heretofore considered, no other forces than the 
loads themselves have been considered. Should forces act 
upon the beam which are not vertical, the force polygon 
will be no longer a straight line, but a broken line somewhat 
similar in character to O l-2-3-Jf-5 in Fig. 321. 



STRENGTH OF MATERIALS. 



785 



In Fig. 332 is shown a cantilever beam projecting 10 ft. 
from the wall. It carries a uniform load of 16 pounds per 
foot of length, and a concentrated load of 40 pounds at a 
distance of 3J feet from the wall. The maximum bending 
moment is required. The method is similar to the last, 
except that, as there is but one support, there can be but 
one reaction. Since the beam is 10 feet long, the total 




Scale of forces 1=100 Ibi. 
Scale of distance 1=3^ 

Fig. 332. 

weight of the uniform load is 16 X 10 = 160 pounds. Hence, 
the reaction = 160 -j- 40 =: 200 pounds. 

Draw O C equal to 200 lb., to some convenient scale. 
Draw P C perpendicular to O C 2it C and choose the pole P 
at a convenient distance from O C. For convenience, 
divide the uniform load into 10 equal parts, as shown; then, 

D. 0. III.— 8 



786 STRENGTH OF MATERIALS. 

each part will represent 16 lb. Lay off O i, 1-2, 2-3, each 
equal to 16 lb,, and 3-Jf equal to 40 lb. Also, Jf-5, 5-6, 6-7, 
etc., equal to 16 lb. each. 3 ft. 3 in. = 3^ ft., and 16x3| 
= 52 lb. Lay off O a = 52 lb., and draw a b, meeting the 
vertical through the center of the weight in b. Draw O b\ 
lay oK b c equal to 40 lb. and draw C n. O b c n is the 
shear line. The perpendicular through the point C coincides 
withP(^; hence, n P C \s> the shear axis. Draw the \mQg ef 
of the moment diagram as in the previous cases. In Fig. 
329, and in the preceding figures, the line g h was drawn 
connecting the extreme ends of the bottom line ; in other 
words, it joined the points where the equilibrium polygon 
cut the lines of direction of the reactions of the supports. 
This cannot be done in this case, because there is no right 
reaction ; therefore, g h must be drawn by means of some 
other property of the polygon. In the previous cases, the 
shear axis was drawn perpendicular to 6^ (7 at the point 
where a line through the pole P parallel to g h cut O C, or, 
in other words, the shear axis was drawn through the point 
which rnarked the end M of the left reaction O M. In the 
present case, Fig. 332, the point C is the end of the left 
reaction; hence, f h is> parallel to P C, /i g is the maximum 
J/, and the bending moment =11 f = P C X h g^= 250 X 45 = 
11,250 inch-pounds. 

It will also be noticed that the distance j = /^ ^ is meas- 
ured from the line ///"upwards, while, for all points between 
the supports in the previous examples, this distance was 
measured downwards. The same observation is true for any 
point between h and f\ hence, for a cantilever, all bending 
moments are negative. 

1392. All the cases of beams heretofore given might 
have been solved by analytical methods — that is, by alge- 
braic processes and formulas ; but the graphical method is 
to be preferred, as it is usually shorter, very nearly as 
accurate, and less liable to error. Moreover, when the dia- 
gram has once been drawn, both the bending moment and 
the shear for any point can be instantly determined. Ex- 



STRENGTH OF MATERIALS. 787 

cept in special cases, the graphical method will be employed 
to determine the maximum bending moment, but in some 
particular cases formulas will be used, by which the result 
can be obtained more quickly than by the graphical method. 



EXAMPLES FOR PRACTICE. 

1. A simple beam 24 feet long carries 4 concentrated loads of 160, 
180, 240, and 120 pounds at distances from the left support of 4, 10, 16, 
and 21 feet, respectively, {a) What are the values of the reactions ? 
{d) What is the maximum bending moment in inch-pounds ? 

Ans \ (""^ ^' "= ^^^^ ^^- ' ^' = ^^^^ ^^• 
* i {b) 28,480 in. -lb. 

2. A simple beam carries a uniform load of 40 pounds per foot, and 
supports two concentrated loads of 500 and 400 pounds at distances 
from the left support of 5 and 12 feet, respectively. The length of the 
beam is 18 feet. What are {a) the reactions ? {b) The maximum bend- 
ing moment in inch-pounds ? * j {a) Ri = 854| lb. ; 7?2 = 765| lb. 

■ ( (b) 48,846 in. -lb. 

3. A cantilever projects 10 feet from a wall and carries a uniform 
load of 60 pounds per foot ; it also supports three concentrated loads 
of 100, 300, and 500 pounds at distances from the wall of 2, 5, and 9 
feet, respectively. Required, (a) the maximum vertical shear, and (b) 
the maximum bending moment in inch-pounds. 

Ans \ (^) - ^'^^^ ^^• 

' ( (b) -110,400 in. -lb. 

4. A beam which overhangs one support sustains six concentrated 
loads of 160 lb. each at distances from the left support of 4 ft. 9 in., 
7 ft., 9 ft. 6 in., 12 ft., 15 ft., and 18 ft. 3 in., respectively, the distance 
between the supports being 16 ft. What are (a) the reactions ? (b) The 

maximum bending nioment ? Ans -i ^'^^ ^^ ~ '^^^ ^^' ' ^^ ~ ^^^ ^^' 

■ i (b) 20,460 in.-lb. 

5. A beam which overhangs both supports equally carries a uni- 
form load of 80 pounds per foot, and has a load of 1,000 pounds in the 
middle, the length of the beam being 15 feet, and the distance between 
the supports 8 feet. What is (a) the vertical shear ? (b) The maximum 
bending moment ? a j,g j («) 820 lb. 

* ] (b) 25,800 in.-lb. 

Note. — The student may not obtain the exact answers given above, 
but if his results do not differ by more than 1%, he will know that his 
method is right. 

NEUTRAL AXIS. 
1393. In Fig. 333, let A B C D represent a cantilever. 
Suppose that a force i^acts upon it at its extremity A. The 
beam will then be bent into the shape shown by A' B CD', 



788 



STRENGTH OF MATERIALS. 



It is evident from the cut that the upper part A' B is 
now longer than it was before the force was applied; i. e., 
^' ^ is longer than A B. It is also evident that U C is 
shorter than D C. Hence, the effect of the force F in bend- 
ing the beam is to lengthen the upper fibers and to shorten 




Fig. 333. 

the lower ones. In other words, when a cantilever is bent 
through the action of a load, the upper fibers are in tension 
and the lower fibers in compression. The reverse is the 
case in a simple beam in which the upper fibers are in com- 
pression and the lower fibers in tension. Further consider- 




FlG. 334. 

ation will show that there must be a fiber, vS S'\ which is 
neither lengthened nor shortened when the beam is bent, 
i. e., S' S" = S S". When the beam is straight the fiber 
5 S", which is neither lengthened nor shortened when the 
beam is bent, is called the neutral line. There may be 



STRENGTH OF MATERIALS. 



78S 



any number of neutral lines dependent only on the width of 
the beam. For, let b a d c, Fig. 333, be a cross-section of 
the beam. Project s upon it in s^. Make b s,^-= a s^ and 
draw s^ s^ ; then, any line in the beam which touches s^ s^ 
and is parallel to 5 S" is a neutral line. Thus, in Fig. 334, 
5j 5', 5 5, S,^ S\ etc., are all neutral lines. The line S^ S^ 
is called the neutral axis, and the surface S^ S' S" S^ is 
called the neutral surface. The neutral axis, then, is the 
line of intersection of a cross-section with the neutral sur- 
face. It is shown in works on mechanics that t/ie neutral 
axis always passes through the center of gravity of the cross- 
section of the beam. 

1 394. Experimental La-w. — When a beam is bent^ the 
horizontal elongation {or compression) of any fiber is directly 
proportional to its distance from the neutral surface^ and^ 
since the strains are directly proportional to the horizontal 
stresses in each fiber ^ they are also directly proportional to 
their distances from the neutral surface^ provided the elastic 
limit is not exceeded. 

1395. Suppose the beam to be a rectangular prism, 
then every cross-section will be a rectangle, and the 
neutral axis will pass through the center o. See Fig. 335. 

Let the perpendicular distance from the neutral axis 
M N to the outermost fiber be denoted by ^, and the 
horizontal unit stress (stress per square inch) at the 
distance c from the axis by S. If a is the area of a fiber, 
the stress on the outermost fibers will be a S. The stress 
on a fiber at the distance unity (1 

inch) from M N is — ; and the stress 

on a fiber at the distance r^ is — X r, 

* c * 



a^ 



The moment of this stress 



about the axis M N is 



aSr, 



X r. = 



aSr^^ 




= — ar' 
c ' 



The moment of the 



Fig. 335. 



790 STRENGTH OF MATERIALS. 

stress on any other fiber at a distance r^ from MN is 

evidently — ^^a', and for a distance r^^ — ■^'^zy ^tc, lin 

is the number of fibers, the sum M of the moments of the 
horizontal stresses on all the fibers is 

5 5 5 

^=y«^' + y^''V + — ^<''+> etc.,= 

<r 5 

—{a r,'+ a r/+ a r^ + a r^) =—a (/-/-j-- r^^^ r,'^ _r„»). 

Now, let r he a quantity whose square equals the mean of 

r ^-\-r ^4- -4-r ^ 

the squares of r^^r^^r^^ r„. Then, r'^^= -5—! — ' ' — - ; 

and, therefore, r^-\- r^ + rg'^ + + '^n— ^^ ^'- Substituting 

above, we get ^ = — •nar'^. But, since a is the area of one 

fiber, nav^ the area of all the fibers — that is, the area A of 
the cross-section ; hence, the sum of the moments of all the 

S 
horizontal stresses = — Ar^, 

c 

1 396. The expression A r", which is found by dividing 
a section into a large number of minute areas (^, a^ etc.), 
multiplying each area by the square of its distance from an 
axis (r^^ r/, r^^ etc.), and then adding the products thus 
obtained, is called the moment of inertia of the section 
with respect to that axis, and is usually denoted by the 
letter /. Hence, 

I=Ar\ (115.) 

1397. The quantity ?', whose square is the mean of 
the squares of all the distances of the minute areas from the 
axis, is called the radius of gyration. 

1398. The sum of the moments of all the horizontal 

SSI 

stresses may then be written as — Ar"^ ^^ — Lor S —. This 

c c c 

expression is called the resisting moment, since it is the 

measure of the resistance of the beam to bending (and, 

consequently, to breaking) when loaded. The resisting 

moment must equal the bending moment when the beam 



STRENGTH OF MATERIALS. 



tn 



is in equilibrium; hence, denoting the bending moment 
by M, 

(116.) 






1399. The values of / and c depend wholly upon the 
size and form of the cross-section of the beam, and S^ is the 
ultimate strength of flexure of the material. 

In Table 29, the average ultimate strength of flexure S^ 
is given for a number of different materials. 

TABLE 29. 



Material. 


Ultimate Strength of 
Flexure in Lb. per Sq. In. 

54. 


Cast Iron 

Wrought Iron 


38,000 
45,000 


Steel 


120,000 


Brass 


17,000 
14,000 


Ash 


Brick 


1,000 


Stone 

Hemlock 


2,000 

7,200 

12,500 


Oak, white 


Pine, white 


9,000 


Pine, yellow 


11,000 


Hickory 


16,000 



1400. Exact values of / for most cross-sections can 
only be determined by the aid of the calculus. The least 
value of / occurs when the axis passes through the center of 
gravity of the cross-section — that is, when / is found with 
reference to the neutral axis. 

The least moments of inertia for a number of different 
sections are given in the Table of Moments of Inertia ; also, 
the area of the sections and the values of c. The dotted 
line indicates the position of the neutral axis, about which 
the moment of inertia is taken. 



792 STRENGTH OF MATERIALS. 

In the Table of Moments of Inertia, A is the area of the 
section, and tt = ratio of the circumference of a circle to its 
diameter = 3.1416. It will be noticed that d is always taken 
vertically. 

1401. To use formula 116, find the bending moment 
in inch-pounds by the graphical method previously described, 
or calculate it by means of the Table of Bending Moments. 
If it is desired to find the size of a beam that will safely 
resist a given bending moment, take S^ from Table 29, Art. 
1 399, and divide it by the proper factor of safety taken 
from Table 28. Then, formula 116 becomes 

M=^, (117.) 

From this— = — q^. Substituting the values of M, f^ and 

S^, the value of ■ — is found. The kind and shape of beam 

having been decided upon, the size can be so proportioned 

that — for the section shall not be less than the value cal- 
c 

culated above. An example will make this clear. 

Example. — What should be the size of an ash girder to resist safely 
a bending moment of 28,000 inch-pounds, the cross-section to be rect- 
angular and the load steady ? 

Solution.— i^=—^ — , or • — — —J—, 
fc c Si 

M= 28,000; from Table 29, Art. 1399, 5* = 14,000; from Table 28, 
Art. 1362,/= 8. 

^^^^>T= 14.000 =^^' 

bd^ d 

From the table of Moments of Inertia, /=— rg— and^ = — for a 

rectangle; hence, — = —-^ x -7 = —a— = 16> o^ b d^ = 96. Any 

number of values of b and <^can be found that will satisfy this equa- 

tion. If b is taken as 6 inches, d^ = -^ = 16 and d= 4/I6 = 4. Hence, 

the beam may be a 6 X 4, with the short side vertical. When possible, 
it is always better to have the longer side vertical. If b is taken as 2 
inches, ^2 _ 43 and d=: |/48 = 7 inches, nearly ; hence, 3 2x7 will also 



STRENGTH OF MATERIALS. 793 

answer the purpose. The advantage of using a 2 X "7 instead of a 6 X 4 

is evident, since the 6x4 contains nearly twice as much material as 

the 2 X '?• Thus, the area of the cross-section of a 6 X 4 is 24 square 

inches, and of a 2 X "7, 14 square inches. Moreover, the 2 X "7, with its 

long side vertical, is slightly stronger than the 6x4, with its short 

/ 2 X ^"■^ 6x4^ 

side vertical, since — = — ^ — = 16J- for the former, and — ^ — = 16 

for the latter. If the 6x4 had its longer side vertical, thus making it 

/ 4 X 6^ 

a 4 X 6, — would then equal — ^j — - = 24, and the safe bending moment 

1^ ,_ . ^ . nyr Si I 14,000X24 ,r,r.r^r.' lu 

could be increased to J/= —? — = ^ = 42,000 m.-lb. 

J c 

1402. If the breaking bending moment, form and size 
of the cross-section of the beam are known, the ultimate 
strength of flexure S^ can be readily found from formula 
1 16, by substituting the values of M^ /, and c^ and solving 
for S,. 

Example. — A cast iron bar, 2 inches square, breaks when the maxi- 
mum bending moment = 63,360 inch -pounds; what is its ultimate 
strength of flexure ? 

Solution. — M= Si — , or 54 = — j^. ^ = 77 = 1'. 

■ ^4 24 , 63,360 X 1 .r; rtOA ^u 

/= ^ = jg- ; therefore, Si = — rTTo^ — = 47,520 lb. per sq. in. 

1403. In order to save time in calculating, the bend- 
ing moments for cases of simple loading are given in the 
Table of Bending Moments. W denotes a concentrated 
load, and w the uniform load per inch of length. All 
dimensions are to be taken in inches when using the 
formulas. 

For any other manner of loading than is described in the 
Table of Bending Moments, the maximum bending moment 
must be found by the graphical method. 

Example. — A wrought iron cantilever, 6 feet long, carries a uniform 
load of 50 pounds per inch. The cross-section of the beam is an equi- 
lateral triangle, with the vertex downwards ; what should be the length 
of a side .' 

Solution. — M=—^ — , from the Table of Bending Moments,— 

50 X (6X12)2 .oft^AA- ^u r bd^ , 2 ,. ^u a> ki 
^ — = 129,600 in. -lb, /= -5^ and c ==-^a, from the Table 

a 00 O 



794 STRENGTH OF MATERIALS. 

/ b d'^ 
of Moments of Inertia; hence, — = -^rr— •5*4 = 45,000, from Table 29, 

Art. 1399, and/= 4, from Table 28, Art. 1362. Therefore, 129,600 
45,000^^^2 129,600 X 4 X 24 __ .^ _. 

= — -, — X -KT-, or d^ = TFT^T^H = 276.48. Since an equi* 

4 24 45,000 

lateral triangle has been specified, b can not be given any convenient 
value in order to find d. For an equilateral triangle, d = b sin 60° = 
.866 b. Hence, bd^z=b (.866 b)'' = .75 b\ Therefore, ^ ^2 _ .75 33 _ 

276.48 or i = ^^^ log i = '"8^76.48 -log. 75 ^ ^^^^^^^ ^^ ^ ^ ^ „._ 
nearly. 

Example. — What weight would be required to break a round steel 
bar 4 inches in diameter, 16 feet long, fixed at both ends and loaded in 
the middle ? 

Solution.— Use formula 116; M=-^. Here i^=— ^— , from 

the Table of Bending Moments, Vol. V; 5* = 120,000; - = — = 

ird^ ^^ Wl ^X (16X12) 120,000 X 3.1416 X 4^ 

-^. Hence, -g- = g = ^ , or W= 

120,000x3.1416x64x8 



16 X 12 X 32 



= 31,416 lb. 



DEFLECTION OF BEAMS. 

1404. The deflection, or amount of bending, produced 
in a beam by one or more loads is given by certain general 
formulas, whose derivation is too complicated to be given 
here. We shall give the formulas only, illustrating their 
application by examples. 

In the third column of the Table of Bending Moments 
are given expressions for the value of the greatest deflection 
of abeam when loaded as shown in the first column. From 
this it is seen that the deflection s equals a constant (de- 
pending upon the manner of loading the beam and upon the 
condition of the ends — whether fixed or free), multiplied by 

-p-j' Let a represent the constant and s the deflection; 

then, 

^ = ^^. (118.) 



STRENGTH OF MATERIALS. 795 

In the above formula, E = coefficient of elasticity, and is 
to be taken from Table 25, Art. 1356, / = length in inches, 
W= concentrated load in pounds, IV' = total uniform load 
in pounds, and / is the moment of inertia about the neutral 
axis; a has values varying from -^ to -J. 

It will be noticed that the deflection is given for only 
nine cases ; for any other manner of loading a beam than 
those here given, it is necessary to use the calculus to obtain 
the deflection. 

Example. — What will be the maximum deflection of a simple 
wooden beam 9 feet long, whose cross-section is an ellipse, having 
axes of 6 inches and 4 inches (short axis vertical), under a concentrated 
load of 1,000 pounds ? 

Solution. — Use formula 118, s = a „ . From the Table of 
Bending Moments, 

^ = -1, IV= 1,000 lb., /= 9 X 13 = 108 in., 

£= 1,500,000 and f=^= """l^^'. 

04. o4 

Hence 1,000 x lOS^ x 64 

iience, ^ - 43 ^ 1,500.000 X ^ x 6 X 64 " 

1405. The principal use of the formula for deflection 
is to determine by its means the stiffness of a beam or shaft. 
In designing machinery, it frequently occurs that a piece 
may be strong enough to sustain the load with perfect safety, 
but the deflection may be more than circumstances will per- 
mit; in this case, the piece must be made larger than is 
really necessary for mere strength. An example of this 
occurs in the case of locomotive guides, and the upper 
guides of a steam engine when the engine runs under. It 
is obvious that they must be very stiff. In such cases it is 
usual to allow a certain deflection, and then proportion the 
piece so that the deflection shall not exceed the amount 
decided upon. 

Example. — The breadtli of the guides of a certain locomotive is not 
to exceed 2J inches. Regarding the guides as fixed at both ends, (a) 
what must be their depth to resist a load of 10,000 pounds at the middle ? 
The guides are made of cast iron and are 38 inches long between the 
points of support, (d) What weight would these guides be able to 
support with safety ? The deflection must not exceed -g^j of an inch. 
The cross-section is, of course, rectangular. 



796 STRENGTH OF MATERIALS. 

Solution. — Since the load comes on two guides, each piece must 

support 10,000-4-2 = 5,000 lb. In the formula, 

1 JV/^ 1 

s= oT^K = ^-Ew-' ^= 192 ^^^ ^^^^ ^^^^' ^=5,000, /=38, 

E= 15,000,000, and /= -^ = %^ = -^ ^'. Substituting, 

1/& Lii lb 

5,000X38^X16 _ 1 
W "^ - 192 X 15,000,000 X 3^^ - 200' 



^r ^^ V5,000X 38^X16X200 ^ ^ g, ^^^^ ^ ^^^^ 

y 192x15,000,000x3 J'' J' -re 

(/5) To find the weight which these guides could support with safety, 

use formula 117, M= ~j, in which J/ = i^, S^ = 38,000,/= 10, 

/ bd^ 2i X (4H)2 50,625 _ , ^.^ ^. 
7 = -6- = r^ = -6:i4i- Substitutmg. 

38,000 X 50,625 X 8 « ..^ ,, . 
^= 10X6,144X38 =^'592 lb. Ans. 

Hence, the beam is over 30^ stronger than necessary, the 
extra depth being required for stiffness. 



EXAMPLES FOR PRACTICE. 

1. How much will a simple wooden beam 16 ft. long, 2 in. wide and 
4 in. deep deflect under a load in the middle of 120 lb. ? Ans. 1.106". 

2. What should be the size of a rectangular yellow pine girder 20 ft. 
long, to sustain a uniformly distributed load of 1,800 lb. ? Assume a 
factor of safety for a varying stress. Ans. 5" X 8". 

3. A hollow cylindrical beam, fixed at both ends, has diameters of 
8 in. and 10 in. If the beam is 30 ft. long and is made of cast iron, {a) 
what steady load will it safely support at 15 ft. from one of the sup- 
ports ? {b) What force will be required to rupture the beam if applied 
at this point ? ^^^ j {a) 8,158 lb. 

'1(^)48,946 lb. 

4. A simple cylindrical wrought-iron beam, resting upon supports 
24 ft. apart, sustains three concentrated loads of 350 lb. each, at dis- 
tances from one of the supports of 5, 12, and 19 ft. ; what should be the 
diameter of the beam to withstand shocks safely ? Ans. 4.71", say 4f". 

5. Find the value of — for a hollow rectangle whose outside dimen- 

c 

sions are 10 in. and 13 in. , and inside dimensions are 8 in. and 10 in. ; 
{a) when the long side is vertical ; (J?) when the short side is vertical. 



A„^U.) 179.103. 
( (*) 1311. 



STRENGTH OF MATERIALS. 797 

6. What is the deflection of a steel bar 1 in, square and 6 ft. long, 
which supports a load of 100 lb. at the center ? Ans. .31104". 

7. Which will be the stronger, a beam whose cross-section is an 
equilateral triangle, one side measuring 15 in., or one whose cross-sec- 
tion is a square, one side measuring 9 in. ? Both beams are of the 
same length. Ans. The one having the square cross-section. 

8. A wooden beam of rectangular cross-section sustains a uniform 
load of 50 lb. per foot. If the beam is 8" X 14" and 16 ft. long, how 
much more will it deflect when the short side is vertical than when 
the long side is vertical ? Ans. .055417". 

COMPARISON OF STRENGTH AND STIFFNESS 

OF BEAMS. 
1406. Consider two rectangular beams, loaded in the 
same manner, having the same lengths and bending 
moments, but different breadths and depths. Then, 

M=S-^ = S^ {l)2indM=S^ (2). Dividing (1) by 

Equation 3 shows that, if both beams have the same 
depth, their strengths will vary directly as their breadths, 
i. e., if the breadths are increased 2, 3, 4, etc., times, their 
strengths will also be increased 2, 3, 4, etc., times. It also 
shows that, if the breadths are the same and the depths are 
increased, the strengths will vary as the square of the depth, 
i. e., if the depths are increased 2, 3, 4, etc., times, the 
strengths will be increased 4, 9, 16, etc., times. Kence, it 
is always best, when possible, to have the long side of abeam 
vertical. If the bending moments are the same, but the 
weights and lengths are different, M = £• IV / (1) and M ^^ g 
Wj /j (2), when ^ denotes the fraction -^, ^, etc., according to 
the manner in which the ends are secured, and the manner 

of loading. Dividing (1) by (2)-^= f"^ ^ , or ^/ = 

W, /, (3). 

Equation 3 shows that if the load W or W^ be increased, 
the length / or /^ must be decreased; consequently, the 
strength of a beam loaded with a given weight varies in- 
versely as its length, i. e., if the load be increased 2, 3, 4, 



798 STRENGTH OF MATERIALS. 

etc., times, the length must be shortened 2, 3, 4, etc., times, 
the breadth and depth remaining the same. 

Example. — If a simple beam, loaded in the middle, has its breadth 
and depth reduced one-half, what proportion of the original load could 
it carry ? 

Solution. — In the preceding paragraphs, it was shown that the 
strength varied as the product of the breadth and the square of the 
depth, or b^dx^ = \y, (i)'^ = i- Consequently, the beam can support 
only \ of the original load. Had the breadth remained the same, 
(^)2 z=z\ oi the original load could have been supported. Had the 
depth remained the same, \ of the original load could have been 
supported. 

Example. — A beam 10 ft. long, loaded in the middle, has a breadth 
of 4 in. and a depth of 6 in. The length is increased to 12 ft., the 
breadth to 6 in., and the depth to 8 in. ; how many times the original 
load can it now support ? 

Solution. — The strength varies directly as the product of the 

b d'^ 
breadth and square of the depth and inversely as the length, or as — j— . 

If ^, d^ and / denote the original sizes, the strength of the two beams will 

, , , ,, b,d,\bd^ 6 X 8^ . 4 X 6\ 6x8^ qo a 

be to each other as — j— : — j— , or as — r^ — • : — ttt — ; — r^— = 32 and 

4X6* 32 

— TjT — = 14.4. , = 2f . Consequently, the beam will support a 

load 2f times as great as the original beam. 

1407. By a process of reasoning similar to that em- 
ployed above, it can be shown that the maximum deflection 
of a beam varies inversely as the cube of the depth and 
directly as the cube of the length. In other words, if the 
depth be increased 2, 3, 4, etc., times, the deflection will be 
decreased 8, 27, 64, etc., times; and, if the length be in- 
creased 2, 3, 4, etc., times, the deflection will be increased 
8, 27, 64, etc., times. Hence, if a beam is required to be very 
stiff, the length should be made as short and the depth as 
great as circumstances will permit. 



COLUMNS. 
1 408. When a piece ten or more times as long as its least 
diameter or side (in general, its least transverse dimension) 
is subjected to compression, it is called a coltimn or pillar. 



STRENGTH OF MATERIALS. 



799 



The ordinary rules for compression do not apply to columns, 
for the reason that when a long piece is loaded beyond a 
certain amount, it buckles and tends to fail by flexure. This 
combination of flexure and compression causes the column 
to break under a load considerably less than that required to 
merely crush the material. It is likewise evident that the 
strength of a column is principally dependent upon its di- 
ameter, since that part having the least thickness is the part 
that buckles, or bends. A column free to turn in any direc- 
tion, having a cross-section of 3" X 8", is not nearly so strong 
as one whose cross-section is 4" X 0". The strength of a very 
long column varies, practically, inversely as the square of 
the length ; i. e. , if a column d is twice as long as a column a, 
the strength oi I? =^ {^Y = ^ the strength of a, the cross- 
sections being equal. 

1409. The conditions of the ends of a column play a 
very important part in deter- 
mining their strength, and must 
always be taken into considera- 
tion. In Fig. 336, are shown 
three classes of columns. The 
column marked a is used in archi- 
tecture, while the columns simi- 
lar to d and c are used in bridge 
and machine construction. 

According to theory, which is 
confirmed by experiment, a col- 
umn having one end flat and the 
other rounded, like d, is 2:^ times fig. 336. 

as strong as a column having both ends rounded, like c. 

One having both ends flat, like a, is 4 times as strong 
as c, which has both ends rounded, the three columns being 
of the same length. If the length of c be taken as 1, the 
length of d may be 1^, and that of a may be 2 for equal 
strength, the cross-sections all being the same; for, since the 
strengths vary inversely as the squares of the lengths, the 

strength of c is to that of <^ as 1 : -7-;-;^^ or as 1 : |. But, 




(HY 



800 



STRENGTH OF MATERIALS. 



since b is 2^ = f times as strong as r, a x f = 1, or, the 
length of b being \\ times that of r, its strength is the same. 
Similarly, when a is tv/ice as long as c its strength is the 
same. 

1410. Columns like b and^ do not actually occur in 
practice, an eye being formed at the end of the column and 
a pin inserted, forming what may be termed a hinged end, 
A steam engine connecting-rod is a good example of a col- 
umn having two hinged ends, and a piston rod of a column 
having one end hinged and one end flat. 

141 1# There are numerous formulas for calculating the 
strength of columns, but the one that gives the most satis- 
factory results for columns of all lengths is the following: 



W^ 



S^A 



A^-fr) 



(119.) 



In this formula, W^= load, S^ = ultimate strength for 

compression, taken from Table 26, Art. 1357,^ = area of 

section of a column in square inches, /= factor of safety, 

/ = length in inches, g = constant, to be taken from Table 

30, and / = least moment of inertia of the cross-section — 

that is, the moment of inertia about an axis passing through 

the center of gravity of the cross-section and parallel to the 

longest side. In other words, if the column has a rectangular 

cross-section, whose longer side is b and shorter side d, the least 

. . b d^ 
moment of inertia is -^r-r-, the axis in this case being parallel 

to the long sides b. The values of g are given in the following 
table : 

TABLE 30. 



Material. 



Timber 

Cast Iron . . . . 
Wrought Iron 
Steel 



Both Ends 
Fixed. 



3,000 

5,000 

36,000 

25,000 



One End 
Hinged. 



1,690 

2,810 

20,250 

14,060 



Both Ends 
Hinged. 



750 
1,250 
9,C00 
6,250 



STRENGTH OF MATERIALS. 801 

Example. — The section of a hollow rectangular cast-iron column 
has the following dimensions (see Table of Moments of Inertia) : 
d=%\ d^ = 6", b = 6", and b^, = 3|". If the length is 10 feet and the 
ends are fixed, what steady load will the column sustain with safety ? 

Solution. — From the Table of Moments of Inertia, least I = 
3-V(^/^3 _ ^^ ^,3)^ ^^» ,0 = 123.5625. A=:bd-b^d, = 6xS 

- 3.5 X 6 = 27 sq. in. ^2 = 90,000, / = 6, / = 10 X 12 = 120", and £■ = 



i^nnn n^u f tj^ 90,000x27 2,430,000 ^,„ ^^^ 

5,000. Therefore, JV = — ^r^z t^^t^^ = -^ToT^f" = 247,800 



27 X 120'^ \ "~ 9.806 
5,000 X 122.5625, 



lb., nearly. Ans. 

Had the column been less than 10 X 6 = GO in. = 5 ft. 

long, the safe load would have been — '- — ^ = 405,000 

lb. Had it been twice as long, it would have supported 

f ^ A f ^ 90,000 X 27 ... ^^^ .. 

a safe load of only— ^^ ^ ^^^^ ^ = 114,500 lb., 

^ V "^ 5,000 X 122.5625/ 
nearly. 

1412. In the actual designing of a column, the size of 

the cross-section is not known, but the form (square, round, 

etc.) is known, also the length, material, condition of ends 

and load it is to carry. To find the size of the cross-section, 

5 
substitute -^ in formula 108, for S, and solve for A^ 

Pf 
obtaining^ ~~^* Substituting in this equation the values 

of /*(= J'F), y, and vS^, this gives the value of A for a short 
piece less than 10 times the length of the shortest side, or 
diameter. Assume a value of A somewhat larger than that 
just found, and dimension a cross-section of the form chosen 
so that its area shall equal that assumed. Calculate the 
moment of inertia and substitute the values of fF, A^ /, /, 

and^ in formula 119, and solve for -j. If the result last 

found equals the value of ~ taken from Tables 26 and 28, 

the assumed dimensions are correct ; if larger, the assumed 
dimensions must be increased ; if smaller, they should be 
D. 0. Ul.—i) 



802 STRENGTH OF MATERIALS. 

diminished, and in both cases the value of -~ should 

be recalculated. An example will serve to illustrate the 
process. 

Example, — What should be the diameter of a steel piston rod 5 feet 
long, the diameter of the piston being 18 inches and the greatest 
pressure 130 pounds per square inch ? 

Solution. — S^ for this case = 150,000 lb. Since the piston rod is 
liable to shocks, a factor of safety of 10 should be used ; hence, -^ = 

i5^^ = 15,000 lb. The load «^= 182 X. 7854x130 = 33,081 lb. A 
Pf 33,081 oo • 1 

Assume that 3 sq. in. are needed. The diameter of a circle corre- 
sponding to an area of 3 sq. in. is y ^^_ ; = 1.9544". Assume the 
diameter to be l^f" = 1.9375; the area will be 1.9375^ x .7854 = 

2.9483 sq. in. The value of /= -^ = aA ^^^^ •;= .69173. W 

^ 64 64 

S.A 



^ ,, ^3 W(. An\ 33,081/, 3.9483 X (5 X 12)2 \ 

Consequently, j =-^[l +-^) =-^^^^ [\ + 14,060 X .69173 ) 

= 23,465 lb. 

As this value exceeds 15,000 lb., the diameter of the rod must 
be increased. Trying 2V as the diameter, the area is 3.5466 sq. in., 

02 QO no-i 

and / = 1.00093. Substituting these values as before, -r- — ^ l^n^ 

J o. o4do 

A 3.5466x602 \ ^„^,^^,^ ^, . • .,, 

I 1 + 14 060 X 1 00093 J ~ ^^'^^^ ^^' This is still too large ; hence, trying 

2i", the area = 3.976 sq. in. /= 1.258 and 

5, _ 33,081 (^ . 3.976 X 6O2 ^ ,.nr;oiK 
7 - "3:976" V + 14,060 X 1.258 j = ^^'^^^ ^^• 
Consequently, the diameter should be 2^". 



EXAMPLES FOR PRACTICE. 

1. What safe steady load will a hollow cylindrical cast-iron column 
support, which is 14 feet long, outside diameter 10 inches, inside diam- 
eter 8 inches, and which has flat ends ? Ans. 273,500 lb. 



STRENGTH OF MATERIALS. 803 

2. A hollow wooden column having a square cross-section is to 
support a steady load of 15,155 pounds. If the thickness of the side is 
1^ inches, length of column 20 feet, and the ends flat, what should be 
the length of the sides of the cross-section, outside and inside ? 

Ans. Outside, 9" ; inside, 6". 

3. Suppose a wrought-iron connecting-rod to have a rectangular 
cross-section of uniform size throughout its length. If the diameter 
of the steam cylinder is 40 inches, steam pressure 110 pounds per square 
inch, and the length of the rod is 12^ feet, what should be the dimen- 
sions of the cross-section of the rod? Ans. 5|" X 9". 

1413. The preceding method for determining the 
dimensions of the cross-section, when the load and length 
are given, is perfectly general, and can, therefore, be used 
in every case. It is, however, somewhat long and cumber- 
some. For the special cases of square, circular, and rect- 
angular columns, the following formulas may be applied, if 
preferred. They seem complicated, but, when substitu- 
tions are made for the quantities given, the formulas will 
be found of relatively easy application. 

For square columns^ the side c of the square is given by 
the formula 



For circular columns^ the diameter d of the circle is given 
by the formula 



^=1.4142 /^'%^^ + /"%^^ (:g^gV^+l^). (121.) 

For rectangular columns^ assume the shorter dimension 
(depth = d). Then the longer dimension (breadth = b) is 
given by the formula 

,^ ^/(l + ^) (122.) 

dS^ 

Should the dimensions given by the last formula be too 
much out of proportion, a new value may be assumed for d, 
and a new value found for b^ 



804 STRENGTH OF MATERIALS. 

Example. — Required the section of a square timber pillar to stand 
a steady load of 20 tons, the length of the column being 30 feet, and 
its ends both flat. 

Solution.— Here 5a = 8,000 lb., / = 8, ^ = 3,000, W = 40,000 lb., 
/ = 30 X 13 = 360 in. These values, substituted in formula 120, give 

_ i/ 40,000^ / 40;000^8 / 40,000 x~8 12 X '6 ^\ ' 
^~^ 2 X 8,000 "^r 8,000 W X 8,000 "^ 3,000 ) 

= 4/20 + i/2l7i36 = 4/20 + 145.35 

= 4/165:35 = 12.90 = 12i\ nearly, or say 13". 

Example. — Let it be required to solve the problem worked out by 
the general method in Art. 1412. 

Solution.— Here ^2 = 150,000 lb.,/= 10, ^ = 14,060, IV = 33,000 lb., 
nearly, and / = 5 X 12 = 60 in. From these data we have 

.3183 TV/ _ .3183 X 33,000 X 10 _ .3183 X 11 _ 

5a ~ 150,000 ~ 5 -'^^^^' 

16 /"■ _ 16 X 6 0-^ _ 8 X 360 _ . __.„ 
~7" - "14;060" - ~T03"~ - ^•"^^^• 

Then, formula 121, 



d = 1.4142 |/.7003 + 4/. 7003 X 4.7970 



= 1.4142 4/. 7003 + 4/3.3593 = 1.4142 4/. 7003 + 1.8328 
= 1.4142 X 1.5916 = 2.25 = 2^% 
as found by the general or trial method. 

The student may apply formula 122 to the solution of 
example 3 in the preceding article. 



TORSION AND SHAFTS. 

1414. When a force is applied to a beam in such a 
manner that it tends to twist it, the stress thus produced is 
termed torsion. In Fig. 337, ^c represents a beam fixed at 
one end; a load W is applied at the end of a lever arm n^ 
which twists the beam. If a straight line ^ ^ is drawn 
parallel to the axis before the load is applied, it will be 
found, after the weight IV has been hung from n, that the 
line c d will take a position ca, forming a spiral. If the load 



STRENGTH OF MATERIALS. 



805 



does not strain the material beyond its elastic limit, c a 
will return to its original position c b when IF is removed. 
It will also be found that the angles a c b and a o b are 
directly proportional to the loads. 




Fig. 337. 

Torsion manifests itself in the case of rotating shafts. 
Instead of one end being fixed, as in the previous case, the 
reisistance which the shaft has to overcome takes the place 
of the force which before was necessary for fixing one end. 
Should the shaft be too small, the resistance will overcome 
the strength of the material and rupture it. 



1415. The angle a o b^ which may be called the angle 
of twist, plays an important part in the designing of shafts. 
For all solid shafts below 11 inches in diameter, the 
following formula may be used: 

d=cYP^=c,\/^, (123.) 

in which d-=^ diameter of round shaft or the side of a square 
shaft in inches; ^ = constant from Table 31; P= force or 
weight applied to the end of the lever arm, in pounds; 
r = length of lever arm in inches, from center of shaft to 
point of application of P; c^ = constant from same table; 
//"= horsepower transmitted, and iV= number of revolu- 
tions per minute. 



806 



STRENGTH OF MATERIALS. 



TABLE 31 





c. 


^.- 


Material. 


Round. 


Square. 


Round. 


Square. 


Wrought Iron 

Cast Iron 


.31 
.353 

.297 


.272 
.309 
.2G 


4.92 
5.59 

4.7 


4.31 

4.89 


Steel 


4.11 







Example. — What should be the diameter of a wrought-iron crank 
shaft for a 16" X 20" steam engine, if the greatest steam pressure is to 
be 90 lb. per sq. in. ? (Assume that the entire steam pressure is trans- 
mitted through the crank-pin at some point of the stroke). 

Solution.— Total pressure on piston = IG^ x .T854 X 90 = 18,095.616. 

20 
say 18,000 lb. = 7^ in formula 1 23. r = —■ = 10'. Therefore. 



A 6|" shaft would be sufficiently large. 

Example. — What horsepower could be safely transmitted by a 7- 
inch cast iron square shaft making 80 revolutions per minute ? 

Solution.— Formula 123 gives, d= c, \/ —, or H= ^^^ = SOxJT^ 

^ y N r,^ 4.89-» 

= 335.93, say 336 horsepower. 

1416. If the diameter of a wrought-iron shaft is 
greater than 12.4 inches, of a cast-iron shaft greater than 
10.3 inches, or of a steel shaft greater than 13. G inches, the 
following formula should be used : 



d=kV'P7=k^ 



(124.) 



k and k^ being taken from Table 32. If the shaft is hollow 
(round), either of the two following formulas may be used: 



or 



H 



'='(^> 
='.-(^> 



(125.) 



(126.) 



d^ and d^ being the outside and inside diameters respectively 
and g and q^ constants to be taken from Table 32. 



STRENGTH OF MATERIALS. 



807 



TABLE 32. 



Material. 


k 


k. 


9 


^1 


Wrought Iron 

Cast Iron 


.0909 
.1145 

.0828 


3.62 
4.56 
3.3 


1,335 

669 

1,767 


.0212 
0106 


Steel 


028 







Example. — What horsepower can be safely transmitted by a hollow 
wrought-iron shaft making 60 revolutions per minute, and whose 
diameters are 9^ and 12 inches ? 

Solution. — 
H^ ^, ^(^^^) = .0212 X 60 (^5^) = 1,384.65 H.P. 

EXAMPLES FOR PRACTICE. 

1. What should be the diameter of a steel shaft to transmit 500 
horsepower at 200 revolutions per minute ? Ans. 5.91", say 5^". 

2. How many horsepower will an 8" round wrought-iron shaft 
transmit with safety, running at 150 R. P. M. ? Ans. 1,048.5 H.P. 

3. A hollow cast-iron shaft has an outside diameter of 10 inches and 
an inside diameter of G inches; at what speed should it be run to trans- 
mit 750 horsepower ? Ans. 81.29 R. P. M. 

4. A wrought-iron shaft 4 inches square runs at 110 revolutions per 
minute; what horsepower will it safely transmit ? Ans. 81.6 H.P. 

5. What should be the diameter of a wrought-iron shaft to transmit 
6,000 horsepower at 100 revolutions per minute ? 

Ans. 14.172", say \^\ 

ROPES. 

1417'. The strength of hemp and manila ropes varies 
greatly, depending not so much upon the material and area 
of cross-section as upon the method of manufacture and the 
amount of twisting. 

Hemp ropes are about 25^ to 30^ stronger than manila 
ropes or tarred hemp ropes. Ropes laid with tar wear bet- 
ter than those laid without tar, but their strength and flexi- 
bility are greatly reduced. For most purposes, the follow- 
ing formula may be used for the safe working load of any of 
the three ropes mentioned above: 

P^\^^C\ (127.) 



SOB STRENGTH OF MATERIALS. 

in which P^= working load in pounds and (7= circumference 
of rope in inches. This formula gives a factor of safety of 
from 7-|- for manila or tarred hemp rope to about 11 for 
best three strand hemp rope. When excessive wear is likely 
to occur, it is better to make the circumference of the rope 
considerably larger than that given by the formula. 

1418. Wire rope is made by twisting a number of 
wires (usually 19) together into a strand and then twisting 
several strands (usually 7) together to form the rope. It is 
very much stronger than hemp rope, and may be much 
smaller in size to carry the same load. 

For iron wire rope of 7 strands, 19 wires to the strand, 
the following formula may be used, the letters having the 
same meaning as in formula 1 27 : 

^=600 6^^ (128.) 

Steel wire ropes should be made of the best quality of 
steel wire ; when so made they are superior to the best iron 
wire ropes. If made from an inferior quality of steel wire, 
the ropes are not as good as the better class of iron wire 
ropes. When substituting steel for iron ropes, the object in 
view should be to gain an increase of wear rather than to 
reduce the size. The following formula may be used in 
computing the size or working strength of the best steel 
wire rope, 7 strands, 19 wires to the strand: 

P= 1,000 (f^ (129.) 

Formulas 1 28 and 1 29 are based ©n a factor of safety of 6. 

1419. When using ropes for the purpose of raising 
loads to a considerable height, the weight of the rope itself 
must also be considered and added to the load. The weight 
of the rope per running foot, for different sizes, may be 
obtained from the manufacturer's catalogue. 

Example. — What should be the allowable working load of an iron 
wire rope whose circumference is 6f inches ? Weight of rope not to 
be considered. 

Solution. — Using formula 128, 

i' = 600 X (6f)2 = 27,337.5 lb. 



'm^ 



.<■■ 



STRENGTH OP MATERIALS. 



809 



Example. — The working load, including weight, of a hemp rope is 
to be 900 pounds ; what should be its circumference ? 
Solution. — Using formula 127, 

^900^3, 

1420» In measuring ropes, the circumference is used 
instead of the diameter, because the ropes are not round 
and the circumference is not equal to 3.1416 times the 
diameter. For three strands the circumference is about 
2.86 <^, for seven strands about 3 d, d being the diameter. 



CHAINS. 

1421. The size of a chain is always specified by giving 
the diameter of the iron from which the link is made. Tlie 
two kinds of chain most gener- 
ally used are the open link 
chain and the stud link chain. 
The former is shown by (a), 
Fig. 338, and the latter by (d). 
The stud prevents the two sides 
of a link from coming together 
when under a heavy pull, and 
thus strengthens the chain. 

It is a good practice to anneal 
old chains which have become 
brittle by overstraining. This 
renders them less liable to snap 
from sudden jerks. The anneal- 
ing process reduces their tensile ^^^' ^^^ 
strength, but increases their toughness and ductility, two qual- 
ities which are sometimes more important than mere strength. 

Let P = safe load in pounds ; 

d = diameter of link in inches. 

Then, for open link chains, made from a good quality of 
wrought iron, 

P= 12,000^', (130.) 

and, for stud link chains, 

P= 18,000^'. (131.) 





I 



810 STRENGTH OF MATERIALS. 

Example. — What load will be safely sustained by a f-inch open link 
chain ? 

Solution. — Using formula 130, 

P = 13,00C d^ = 12,000 X (f )* = 6,750 lb. 

Example. — What must be the diameter of a stud link chain to carry 
a load of 28,125 pounds ? 

Solution. — Using formula 131, /*= 18,000 </*. Hence, 



._ ./ZZI _ i./ ^8,125 _ 
y 18,000" r 18,000" *• 

1421^. The statement in the last sentence of Art. 
1358 may be modified in practice to include pieces whose 
lengths are not greater than ten times their least transverse 
dimensions, without material error. Although the stress is 
pure compression only for pieces whose lengths do not exceed 
Jive times their least transverse dimensions, the results 
obtained by formula 119 agree so closely with those ob- 
tained by formula llO^when the length does not exceed 
ten times the least transverse dimension, that the latter 
may be used in all such cases. When the length is greater 
than ten times the least transverse dimension, the piece 
becomes a column, and formula 119 must be used. 



APPLIED MECHANICS. 



1422. A mactLiiie is an assemblage of moving parts.^ 
together with a supporting frame so arranged as to utilize 
some external source of energy for the purpose of doing 
work. 

In the operation of machinery, motion and force are com- 
municated to one of the moving parts, and transmitted to 
the part where the work is done. During the transmission , 
both the motion and force are modified in direction and 
amount, so as to be rendered suitable for the purpose to 
which they are to be applied. 

The moving parts are arranged to have certain definite 
motions relative to each other, the effect of which is to com- 
pel the piece where the Avork is done to have the required 
motion. The nature of these movements is independent of 
the amount of force transmitted ; in other words, in a model 
of a machine, operated by hand, the relative motions of the 
parts will be precisely the same as in the machine itself, 
although, in the latter case, a great amount of power may 
be transmitted and much work done. 

14:!23. Kinematics is that branch of Applied Mechan- 
ics which treats of the motions of the parts of a machine, 
without regard to the forces acting. 

1424. The dynamics of machinery is that branch 
which treats of the forces acting in the operation of 
machinery. The dynamics of machinery depends upon the 
subject of Kinematics, since every change in force in a 
machine is the result of a change in motion. Thus, in the 
steam engine, if the motion of the piston be transmitted and 
modified so as to cause the periphery of the fly-wheel to 
move eight times as fast as the piston, the force exerted 
upon the belt would be only one-eighth of the steam pressure 
upon the piston. 

For notice of copyright, see page immediately following the title page. 



812 APPLIED MECHANICS. 

In what follows, Kinematics will be treated of mainly, 
though in some cases the forces acting will be considered. 

1425. Medianism is a term applied to three or more 
parts of a machine, so combined that the motion of the first 
compels the motion of the other movable parts, according to 
the law depending upon the nature of the combination. 

The terms mechanical movement and mechanical motion are 
often used as having the same meaning as mechanism. A 
machine is made up of a number of mechanisms. 

1426. Driver and Follo^wer. — That piece of a mech- 
anism which causes motion is called the driver, and the 
one whose motion is effected is called the follo-wer. In the 
case of belt or toothed gearing, the follower is often called 
the driven wheel. 

1427. Right-tianded rotation is rotation in the 
direction of the motion of the hands of a watch. Left- 
lianded rotation is in the opposite direction Looking at 
a rotating pulley from one side, its rotation would be right- 
handed, if it turned in the same direction as the hands of a 
watch, held and looked at by the observer. Viewing the 
pulley from the other side, its rotation would be left-handed. 

1428. Cycle of Motions. — When a mechanism is set 
in motion, and its parts go through a series of movements, 
which are repeated over and over in the same order, each 
series is called a cycle of motions. 

1429. Velocity ratio is a term used to signify the 
comparative velocities of two pieces. Thus, if two gear 
wheels are so proportioned that one turns three times as 
fast as the other, their velocity ratio would be 3 or \^ accord- 
ing as the more rapidly revolving gear was mentioned first 
or last. 

LINK MECHANISMS. 

1430. A bar, or other rigid body, connecting two ele- 
ments or parts of a mechanism, is termed a link. In the 
steam engine, the crank and connecting-rod are links, and 
the engine frame may be considered to be a third or closing 



APPLIED MECHANICS. 



813 




:l^ 



link, one end of which supports the crank-shaft, the other 
end being slotted to guide the cross-head. 

Links have special names, according to the machine or 
location in which they are used. Thus, a link which 
vibrates about a point is called a beam, rocker, or lever, 
and one which turns completely around a point is called a 
crank. A link connecting with an oscillating, or rotating, 
link is called by various names, as connecting-rod, crank- 
rod, pitman, eccentric-rod, coupling-rod, parallel- 
rod, etc. In practice, the word "link" is applied mainly 
to slotted links, such as the link in "Stephenson's link 
motion." 

In what follows, when speaking of the length of a link, 
the distance be- 
tween centers will 
be understood, or 
the distance A^ in 

Fig. 339. Fig. 339. 

1431. The center line of motion of any mechanism 
is a straight line so drawn as to represent the general or 
mean direction in which one or more of the parts move. 
When the motion of the parts is not in a straight line, their 
deviation occurs equally on each side of this center line. 
Thus, in the steam engine, the center line of motion runs 
from the center of the cylinder to the center of the shaft, 
the connecting-rod vibrating equally on each side of the line. 

1432. Levers. — Levers are used in mechanisms to 
guide a moving point, as the end of a moving rod, or to 
transfer motion from one line to another. There are three 
cases: (I) Levers whose lines of motion are parallel; (II) 
levers whose lines of motion intersect, and (III) levers 
having arms whose center lines do not lie in the same plane. 

In proportioning levers, the following points should in 
general be observed. They apply to all three cases just 
mentioned : 

(1.) When in mid position, the center lines of the arms 
should be perpendicular to the lines along which they give 



814 



APPLIED MECHANICS. 



or take their motions, so that the lever will vibrate equally 
each way. 

(2.) If a vibrating link is connected to the lever, its point 
of attachment should be so located as to move equally on 
each side of the center line of motion of the link. 

(3.) The lengths of the lever arms must be proportional 
to the distances through which they are to vibrate. 

1433. Case I. — Reversing Levers. — An example of 
a lever which illustrates the foregoing principles is shown 
in Fig. 340. The crank R S is the driver, and gives a 




Fig. 340. 

motion along the center line A B^ which is transferred by 
the lever E H to the line C D. The lever vibrates equally 
each way about its fulcrum or center (9, as indicated by the 
lines c b and d a. When in mid position, its center line 
E H IS perp endicular to the lines of motion C D and A B, 
The horizontal distances traversed by points E and H^ re- 
spectively, are proportional to the arms E O and H O^ or 
Y : E O = X : H O. The vibrating link E K connects 
the point E with the rod K D^ which is constrained to move 
in a straight line by the guide G^ and in accordance with 
principle 2, the lever is so proportioned that point E will be 
as far above the center line of motion C D, when in mid 
position, as it will be below it in the two extreme positions; 



APPLIED MECHANICS. 



815 



that is, points c and d are as far below the line as point E is 
above it. 

At the bottom of the lever, where the rod H R connects 
with the crank, the same principle holds, point H being as 
far below the line A B sls points a and d are above it. 

Frequently, the distance between the center lines C D and 
A B is given, and the extent of the motion along these lines, 
from which to proportion the lever. A correct solution to 
this problem is troublesome by calculation, because it is not 
known at the start how far above and below their respective 
lines of motion the points £ and 1/ should be. 

1434. It may 

easily be done graph- 
ically, however, as 
shown in Fig. 341. 
Draw the center lines 
of motion C D and 
A B and a center line 
5" T perpendicular to 
them. Draw ME par- 
allel to 5 7" at a dis- 
tance from it equal to 
\ F, or half the stroke 
along C D\ also, the 
parallel line H N^ on 
the other side of vS" Z", 
and at a distance from B. 



it equal to \ X, or half I 



A 




the stroke along A B. N 

Connect points M and 

iV by a straight line; y 

where this line inter- fig. ui. 

sects vS 7", as at O, will be the center or fulcrum of the lever. 

With (9 as a center, find by trial the radius of an arc which 

will cut 5 r as far below the line ^ ^ as it does H N above 

this line, or so that the distance m will equal the distance n. 

As an aid in determining the correct radius, describe an arc 

cutting 5 r, with as a center and a radius O N, The 



816 



APPLIED MECHANICS. 



distance n will be a little more than -J /. Now, draw a 
straight line through points H and O. The part included 
between H N and M E determines the length of the lever. 
In this case, the length of the shorter arm is equal to O E^ 
and of the longer arm, O H. 

1435. Non-Reversing Levers. — Fig. 342 shows the 
same construction applied 



to a lever in which the 
center O is at one end of 
the lever. This lever 
does not reverse the mo- 
tion like the preyious one, 
since when the motion 
along A B is to the right 
or left, the motion along 
C D will be in the same 
direction. The figure is 
lettered like the preceding 
one, so that the construc- 
tion will be easily under- 
stood. 




Fig. 342. 



1436. It is often desirable that a lever mechanism shall 

reproduce upon a greater or 
smaller scale, along one line, the 
exact motion that occurs along 
another line ; that is, that every 
change in the rate of the motion 
along one line shall reproduce a 
corresponding change along the 
other line. Figs. 343 and 344 
illustrate three indicator-reduc- 
ing motions that accomplish this. 
In Fig. 343 the lower end of 
the lever attaches to the cross- 
Z^ head of the engine through the 
swinging link H R. The indi- 
cator string is fastened to the 
bar C D^ which receives its motion from the lever through 




APPLIED MECHANICS. 



817 



the link E A", and slides through the guides G^ G^ in a direc- 
tion parallel to the line of motion A B oi the cross-head. 
In order that the bar C D shall have the same kind of motion 




Fig. 344. 

as the cross-head, it is only necessary that the links E R 
and H R shall be proportional to their respective lever arms; 
thus O H \ H R= O E \ E K. The pins must be so placed 
that the connecting links will be parallel ; if parallel at one 
point of the stroke, they will be so at all points. 

1437. It is to be observed that the pins (9, K^ and R are 

in one straight line, and, in general, it may be said that any 
arrangement of the lever which will keep these three pins in a 
straight line for all points of the stroke will be a correct one. 
In Fig, 344 are two such arrangements. In the first, the 
pins K and R are fast to the slide and cross-head, respectively, 
and slide in slots in the lever. In the second, they are fast to 
the lever, the slots being in the cross-head and slide. In 
both, the pins K and R are in a straight line with the pin O 
during the whole stroke. 

1438. Case II. — Bell-Crank Levers. — Levers whose 
lines of motion intersect are termed bell-crank levers. 

Z>. 0. lii.—io 



818 APPLIED MECHANICS. 

In Figo 345, suppose the angle CAB, made by the lines 



d w 




of motion, to be given, and that the motion along A B is to 
be twice that along^ A C. Draw c d parallel to A C fit any 




Fig. 346 



APPLIED MECHANICS. 



819 



convenient distance from ^ C. Draw ^ 3 parallel to A B 
and at a distance from A B equal to twice the distance of 
c d from A C. Through the intersection of these two lines 
and the apex A of the angle, draw the line A F. Then the 
center O of the bell-crank may be taken at any point on 
A F suited to the design of the machine. Having chosen 
point O, draw the perpendiculars O E and O //, which will 
be the center lines of the lever arms. 

1439. In Fig. 346 a construction is shown that may be 
employed when the two lines C D and A B do not intersect 




Fig. 347. 



within the limits of the drawing. In Fig. 347 the same con- 
struction is applied to a non-reversing lever, in which the 
center O falls outside of the lines A B and C£>. The figures 



820 



APPLIED MECHANICS. 



are lettered alike, and the following explanation applies 
to either: Draw c d parallel to C D^ and a b parallel to 
A B^ as before, so that the distance oi c d from C D : dis- 
tance oi a b from A B = amount of motion along C D : 

^_^ amount of motion along 

4— V^A ^ ^* ^R^ii^» draw lines 

* ' ^^ ^ g h and ^y in exactly the 

same way, but taking care 
to get their distances from 
C D and A B different 
from those of the lines just 
drawn. Thus, if ^ </ should 
be six inches from C D, 
make g h some other dis- 
tance, as four inches, or 
eight inches, and then 
^^°- ^^- draw ^ y at a proportion- 

ate distance from A B. Through the intersections oi a b 
with c d^ and oi e f with g h, draw the line / F, which 
will be the line of centers for the fulcrum O. 

1440. Case III. — Levers falling under this case 
are usually bell-crank levers, with their arms separated by 
a long hub, so as to lie in different planes. They introduce 
no new principle. See Fig. 348. 

1441. Crank and Connecting-Rod. — Fig. 349 is a 
diagram of the crank mechanism used in steam engines, 

dt 





Fig. 349. 

power pumps, etc. A C represents the stroke of the cross- 
head, O the center of the crank-shaft, and a d^c d,^ the path 
described by the crank-pin, called the crank-pin circle. 



APPLIED MECHANICS. 821 

In the crank motion, the relative positions of the cross- 
head and crank-pin vary at every point of the stroke. This 
irregularity, which has an important influence in the design 
of steam-engine valve gears, may readily be observed by 
plotting a few points of the motion. Having drawn the 
crank-pin circle and the center line of motion A <r, set the 
compasses to a radius equal to the length of the connecting- 
rod, which, in this case, is three times the length of the 
crank, or 3 X O a. With ^ as a center, strike an arc 
at A^ and with r as a center, an arc 2X C. AC equals 
the length of the stroke of the cross-head, A and C being 
the extreme positions of the stroke. The corresponding 
positions a and c of the crank-pin are known as the dead 
points or dead centers, because the crank can not be 
started at these points by a direct pressure on the con- 
necting-rod. 

With some point near a on the crank-pin circle, as b^ for a 
center, and with the same radius as before, strike arc B on 
the stroke line. Mark point e on the crank-pin circle, so 
that arc e c— b a\ with ^ as a center, and with the same 
radius, strike arc E. It will be seen that the distance E C 
is less than A B, which shows that the relative motion during 
the first half stroke is different from that during the second 
half. The reason for this is that one-half the crank-pin 
circle curves towards the cross-head, and the other half away 
from it. 

The greatest irregularity occurs when the crank is in its 
middle positions, or at points d^ and d^. With either point 
as a center, strike an arc, as before, which will fall at D^ a 
distance equal to M D from the mid-stroke position of the 
cross-head. Suppose, the crank to turn in the direction of 
the arrow, the cross-head will have passed mid position J/, 
when the crank-pin reaches its middle point at d^\ on the 
return stroke the reverse will be true, the crank-pin reach- 
ing point d^ before the . cross-head reaches M. Hence, 
during the forward stroke, the cross-head moves ahead of 
the crank; on the return stroke, the cross-head lags behind 
the crank. 



822 



APPLIED MECHANICS. 



The common way of plotting the motion is as follows: 
Take for example, the point d^. With d^ as a center, strike 
the arc at D^ as before ; with Z^ as a center and same radius, 
describe arc d^ d^. O d^ is the displacement of the cross- 
head from mid stroke. Point f was obtained in the same 
way by increasing the length of the connecting-rod to F d^^ 
showi?tg that the longer the connecting-rod^ the less the 
irregularity. 

1442. Crank and Slotted Cross-Head. — If the con- 
necting-rod in Fig. 349 be increased to a very great length, 
an arc drawn through d^^ corresponding to the arcs d^ d^^ and 
^xfi would be nearly a straight line coinciding with d^ Oy 
and the horizontal movement of the crank-pin would, there- 
fore, be practically the same as that of the cross-head. If 
the connecting-rod were increased to an infinite length, the 
two movements would be exactly the same. Fig. 350 shows 
the crank and slotted cross-head mechanism by which this 



/^ 



-F 




Fig. 350. 

is accomplished. Consider the crank O B 3.S the driver. 
The crank-pin .^ is a working fit in the block R which is 
arranged to slide in the slotted link L. The rods F and H 
are rigidly attached to the link, and are compelled to move 
in a straight line by the guides G, G'. As the crank 
revolves, the rods F and H are given a horizontal motion 
exactly equal to the horizontal motion of the pin B. 

This mechanism is often applied to steam pumps, where 
one of the rods, as F^ is the steam piston rod, and the other 



APPLIED MECHANICS. 



823 



is the plunger rod. A fly-wheel is driven by means of the 
slotted link and crank, and its kinetic energy makes it pos- 
sible to cut off the steam before the end of the stroke. 
Without the fly-wheel full steam pressure must be carried 
throughout the stroke. 

1443. In Fig. 350, if the crank rotates uniformly, the 
motion of the sliding rods is said to be harmonic, and the 
mechanism itself is often called the harmoniC'-iiiotion 
mechanism. Harmonic motion may be defined as the 
motion executed by the foot of a perpendicular let fall on 
the diameter of a circle from a point moving with uniform 
velocity along the circumference. 

1444. Slow-Motion Mechanism. — A mechanism 
consisting of two connected levers, or of a crank and lever, 
can be proportioned to produce a slow motion of one of the 
levers. 

Such a combination is shown in Fig. 351, where two levers, 
A and B, are arranged to turn on fixed centers, and are 
connected by the rod R. Lever A is actuated by the handle 
i7, secured to the same shaft. If H and lever A be turned 
left - handed, lever B will turn right - handed, but with a 
decreasing velocity, which will become zero when the lever 
A reaches position A^ in line with the rod, which will then 
be in position R^. Any further motion of A will cause B to 
return towards its first position, its motion being slow at 
first and then faster. The mechanism, it will be observed, 
is proportioned contrary to the principles stated in Art. 
1432, and produces a motion that is variable, but very 
powerful. 

To obtain the greatest advantage, the lever B should be 
so placed that it will occupy a position perpendicular to the 
link R at the instant when A and R are in line. To lay out 
the motion, therefore, supposing the positions of the centers 
and lengths of the levers to be known, describe arc I? a about 
the center (9, with a radius equal to the length of lever B. 
Through C^ the center of lever A^ draw the line M N tan- 
gent to the arc just drawn. R and A must then be in line 



824 



APPLIED MECHANICS. 



along the line M N^ and lever B must be perpendicular to 
It when in position B^. Generally, there will be a certain 
required amount of movement for lever B. To secure this, 
draw B in its extreme left-hand position ; then with C as a 
center, and a radius equal to the length of A^ strike the arc 




Fig. 351. 

dc. Set the compasses to length a d^ and from ^ as a center 
strike an arc cutting the arc c d, C c will be the second 
position of lever A, 

In the above combination, if the parts were proportioned 
to allow A to rotate like a crank, R and A would come into 
line twice during each revolution. 

This mechanism has been used to operate platen printing 
presses, where oscillation of the handle H moves the platen 
to and from the type, through the lever B. It is also used 
to operate the exhaust valves in Corliss engines, in a manner 
which will be explained later. 

1445. The toggle-joint, shown in Fig. 352, is a modi- 
fication of the foregoing mechanism, and is much used in 
presses, punching and shearing machinery, etc. It consists 
of the two links O R and C R^ point O being fixed and point 



APPLIED MECHANICS. 



825 



C constrained to move in a horizontal line. A considerable 
movement of point R produces a very small movement in C 
d 







F 

f 


^ 








i 




ik 


\^/:J^ 




p 


sA 


v^c 













Fig. 352. 
and consequently a pressure /^, applied at R, can be made 
to produce a powerful pressure, or thrust, in a direction 
opposite to that of the arrow P. 



THE FORCES ACTING IN LINK MECHANISMS. 

1446. In order to understand how the forces act in a 
link mechanism, it will be necessary to again refer to the 
subject of moments, which is treated of in Elementary 
Mechanics. 

The moment of a force about any point is the product 
of the Tuagnitude of the force and the perpendicular distance 
from the point to the line of action of the force. 

The tendency of a force to rotate a body about a point is 
measured by the moment of the force about that point. For 
example, let the crank C^ in Fig. 353, be 
pivoted at O^ and suppose a force P to act 
upon the crank-pin in the direction shown. 
Then, if the perpendicular distance from 
O to the line of action of the force be O Ay 
the moment of the force tending to rotate 
the crank about O \?> P Y^ O A. If the 
force be stated in pounds and the perpendic- 
ular distance in inches, the product will 
be in inch-pounds; if the perpendicular distance be given in 




Fig. 35d. 



826 



APPLIED MECHANICS. 



feet, the product will be in foot-pounds. These expres- 
sions, however, bear no relation to foot-pounds of work^ and 
the student must avoid confusing the two. 

In the case of a shaft having a pulley, gear, crank, or lever 
attached to it upon which a force acts tending to cause 
rotation of the shaft, the moment of the force is generally 
called the tivisting moment. 

1447". If two or more forces in one plane act upon a 
body, and are in equilibrium, then the sum of the moments 
which tend to turn the body in one direction about a point is 
equal to the sum of the moments of the forces which tend to 
turn the body in the opposite direction about the same point. 
Or, to state the principle more concisely, the opposing mo- 
ments about the point are equal. This is called the princi- 
ple of moments. 

In the crank and connecting-rod mechanism, shown in 
outline in Fig. 354, the tendency of the force P to cause 




Fig. 854. 

rotation of the crank about point O may be determined by 
resolving this force into two forces, one of which acts along 
the connecting-rod in the direction of the arrow, and which 
we will call P\ and the other of which acts in a direction 
perpendicular to the guides. Draw O e perpendicular to 
C R produced; then, assuming the parts to be stationary 
for the instant, so that the effects of inertia may be neglected. 



APPLIED MECHANICS. 



827 



the twisting moment = P' X O e. A simpler method, how- 
ever, is the following: Let the force P act in the direction 
of the line O C, From O draw O A perpendicular to 
O C, and note the point d where it intersects the center 
line of the connecting-rod, or of the center line produced. 
Then, it can be proved that the twisting moment about O 
due to the force Pis Px O d. 

If a belt pulley B is attached to the shaft, the force P 
will be resisted by the pull F of the belt; and, by Wi^ prin- 
ciple of viome7its^ F X O r = Px O d; whence, 

^><^^ (132.) 



F = 



Or 



Example. — In a power pump, if a belt pull of 120 pounds is exerted 
upon the rim of the driving pulley, 24 inches in diameter, and the 
crank is in such a position that the distance O d (Fig. 354) = 2 inches, 
what is the water pressure per square inch upon the pump plunger, if 
its diameter is 4 inches ? 

Solution. — Radius of pulley = 12" = (9 r. From formula 132, 



120 = 



12 ' 



whence P = 720 lb. Area of piston = 12.57 sq. in. 
720 -r- 12.57 = 57.28 lb. per sq. in. Ans. 

1448. A drawing in which the arms, rods, and links of 
a mechanism are indicated by their center lines only is called 
a skeleton diagram. 

Fig. 355 is a skeleton 
diagram of the motion 
shown in Fig. 351, the 
different parts being in 
the same position. Draw 
the line C s perpendic- 
ular to d Cy and suppose a 
force F to act at the end 
of the lever C h. By 
the principle of moments, 
the pull along b c would then be such that F X C h =■ pidj 
oxib cxC s, ox 

'^. (133.) 




Fig. 355. 



Pull on ^ ^ = 



828 



APPLIED MECHANICS. 



The force necessary to balance the pull on b c would be 
supplied by the resistance to motion by whatever force 
might be applied to the shaft O or the lever b (9, and would 
be felt at the point b. If it were desired to find the twist- 
ing moment about the point O^ we would multiply the pull 
on ^ <: by the perpendicular distance O e. 

Example.— In Fig. 355, if F- 25 lb., C// = 2 ft., C.r = yV". and Oe = 
3", {a) what would be the pull at 3 ? {p) What is the twisting moment 
about O ? 

9Fi N/ 9.4. 

6,000 lb. Ans. 



Solution.— (^) Pull = ^ = 25 x 24 X 10 



tV 



{b) Twisting moment = 6,000 X 3 = 18,000 inch-pounds. Ans. 

1449. We will now consider the forces acting in the 
toggle-joint, In Fig. 356, let a pressure F be* exerted upon 




Fig. 356. 

the handle in a direction at right angles to O r. The twist- 
ing moment about O is F X O r^ and the case becomes ex- 
actly similar to that of the connecting-rod crank, O R corre- 
sponding to the crank and C to the cross-head. From O 
erect the perpendicular which intersects the center line of the 
link C R^ extended, at d. Then, as before, Px O d= Fx 
O r. It will here be more convenient to have the formula 
in terms of F. Hence, 



APPLIED MECHANICS. 



829 



The same formula applies when the force -F acts as in Fig. 
352. Drawing the line O r perpendicular to the line of 
action of i% we have the twisting moment F X O r resisted 
hy Px O d. 

1450. When the two links O R and R Care equal in 
length, the height h of the point R above a straight line 
drawn through points O and C will equal \ O d. Hence, 
for equal links, formula 1 34 may be written, 

Fx Or 



P = 



2 A 



(135.) 



Example.— In Fig. 18, if 6>r = 30", Od=Q\ and 7^=100 lb., what 
thrust would be produced by the block C ? 
100 X 30 



Solution. — P = 



6 



= 500 lb. Ans. 



EXAMPLES FOR PRACTICE. 

1. In Fig. 355, let F= 250 lb., CA = 2 it, Cs = 2 in., anddO = 10 
in. If, in this position, the lever dO is perpendicular to d c, what force 
would have to be exerted at a point midway between d and O, in order 
to resist the force F7 Ans. 6,000 lb. 

3. If the piston of an engine is 6 inches in diameter and the steam 
pressure 45 lb. per sq. in., what would be the tangential pressure at 
point i?, in Fig. 354, when the crank is in such a position that Od = 
3 in., the length of the crank being 8 in. ? Ans. 477.13 lb. 

3. What thrust would be exerted by the block C, in Fig. 356, if the 
force i^were to act in a vertical instead of a slanting direction ? Take 
J^= 100 lb. , O r = 60 in. , Od=l in. , and distance of point r above the 
line OC=Sm. Ans. 5, 946i lb. , nearly. 

QUICK-RETURIV MOTIONS. 

1451. Quick-return motions are used in shapers, Blot- 
ters and other machines, where all the useful work is done 
during the stroke of a reciprocating piece in one direction. 
During the working stroke the tool must move at a suitable 
cutting speed, while on the return stroke, when no work is 
performed, it is desirable that it should travel as rapidly as 
possible. 

Vibrating-Link Motion. — The mechanism shown in 
Fig. 357 has been applied to shaping machines operating on 



tl A 



830 



APPLIED MECHANICS. 



metal. Motion is received from the pinion P^ which drives 
the gear G. The pin b is fast to the gear, and pivoted to it 
is the block B^ which is fitted to the slot of the link C D. 
As the gear rotates, the pin describes the circle b e d c^ the 
block sliding in the slot of the link C D^ causing C D\.o oscil- 

mi 




late about the point D^ as indicated by the line C C in the 
figure. The rod L connects the upper end of the link with 
the tool slide, or "ram,"i?, which, therefore, oscillates with 
it, but is constrained by guides (not shown) to move in a 
straight, horizontal line. 



APPLIED MECHANICS. 



831 



During the cutting stroke, tlie pin b travels over the arc 
d c b^ or around the greater arc included between the points 
of tangency of the center lines C D and C D. During the 
return stroke the pin passes over the shorter arc bed, 
and as the wheel G revolves with a uniform velocity, the 
lines of the forward and return strokes will be to each 
other as the length of the arc d c b is to the length of the 
arc bed. The throw of the slotted link and the travel of 
the tool can be varied by the screw j, which moves the 
block B to and from the center of the gear. The rod Z, 
instead of vibrating equally above and below a center line 
of motion, is so arranged that the force moving the ram 
during the cutting stroke will always be downwards, caus- 
ing it to rest firmly on the guide. 



1452. To lay out the motion, proceed as follows : Draw 
the center line 5 7", Fig. 358, and parallel to it the line m n, 
the distance between the two being equal to one-half the 
longest stroke of the tool. About (9, which is assumed to be 
the center of the gear, describe the circle b d c with a radius 
equal to the distance from the center of 6^ to <^ (Fig- 357) 
when set for the longest stroke. Divide the circum- 
ference of the circle into an upper and lower arc, extending 
equally on each side of the center line, and having the same 
ratio as the forward and backward strokes. In this case 
the return is 2 to 1, and the circle is divided into three equal 
parts, as shown at b, d, and c, thus making the arc d b equal 
to one-half deb. Draw the radial lines O b and O d. 
Through b draw CD, perpendicular to O b\ the point Z>, 
where it intersects vS T, will be the fulcrum, and the point C, 
where it intersects m n, the upper end of the slotted lever. 
Through (7 draw the horizontal line C C\ making C E equal 
to C E. Draw O D, which should be tangent to the circle 
at d^ thus giving the other extreme position of the lever. 
It is to be observed that for a quick return of 2 to 1, the only 
condition is that lines C D and C D shall be perpendicular 
to O b and O d^ respectively; points 6^ and /^and the length 
of radius O b can be varied considerably. 



832 



APPLIED MECHANICS. 



1453. To plot the motion, draw the center line of 
motion R L through the point at which the connecting-rod 
attaches to the tool slide. Divide the circle deb into a 
number of equal parts, as 7, ^, 3^ etc., and from D draw 
lines through these points, extending to the arc C C. 
Number the points of division on the circle, and give corre- 
sponding numbers to the points of intersection on the arc C C. 
With these last points as centers, and with a radius equal to 
the length of the connecting-rod Z, in Fig. 357, strike arcs 




Fig. 358. ^ 

cutting the line R L, and number these intersections so that 
they will correspond to the other points. In Fig. 358, sup- 
posing the gear to turn with a uniform motion, the tool slide 
will move along i?Z from point 1 to 2 during the first -^^ of a 
revolution ; during the next -^^ revolution, from point 2 to 8, 
etc., on the forward stroke. On the return stroke, from 
point 9 to 10, 11, 12, and 1, the motion is much less uniform. 
Another point about this motion is that, as the radius O b 
is diminished to shorten the stroke, the return becomes less 



I 



APPLIED MECHANICS. 



833 



rapid, as can be seen from the dotted lines in the figure, 
which show the radius O b shortened to O b\ and the corre- 
sponding position of C D. 

1454. Whitworth Quick-Return Motion. — This 
mechanism is shown in principle in Fig. 359. The pin b^ 
inserted in the side of the gear (7, gives motion to the slotted 
link C D^ as in the vibrating link motion. This motion 
closely resembles the previous one, the difference being that 
the center D^ of the slotted link, lies within the circle described 
by the pin b^ while in the previous case it lies without it. 
To accomplish this result, a pin P is provided for the gear 
to turn upon, and is made large enough to include another 
pin D^ placed eccentrically within it, which acts as the center 
for C D. With this arrangement, the slotted link follows 
the crank-pin during the complete revolution, instead of 

8 




vibrating, and thus becomes a crank. The stroke line R L 
passes through the center of D^ which is below the center of 
P. The forward, or working, stroke occurs while the crank- 
pin b passes over the arc d c g^ and the quick return occurs 
while it travels over the arc g e d. During each of these 

i>. 0. 111.— 11 



834 



APPLIED MECHANICS. 



intervals, the link C D completes a half revolution, and, con- 
sequently, must move more rapidly, while the crank-pin 
describes the shorter arc. 

1455. To proportion the motion, it is only necessary to 
so locate the stroke line R L that it will divide the crank-pin 
circle d c g e into two parts, d c g and ^ ^ ^, in the propor- 
tions of the forward and return strokes. The point D^ where 
this line cuts the center line 6" 7", is the position for the 
center of the slotted crank. The motion is plotted as in 
Fig. 360. Divide the crank-pin circle d c g e into a number 

5 




Fig. 360. 

of equal parts. From D^ the center of the slotted crank, 
draw radial lines through these points to the outer circle, 
which represents the path of pin C (Fig. 359), using the latter 
points of intersections as centers, and with a radius equal to 
the length of the connecting-rod, strike off points on the 
stroke line, which will show the movement of the tool for 
equal amounts of rotation of the driving gear. 

1456. Fig. 361 shows the mechanism as practically 
constructed. The gear G is driven with a uniform velocity, 
in the direction of the arrow, by the pinion H. It rotates 
upon the large pin P (which is a part of the frame of the 
machine), and carries the pin b which turns in the block k 
and is capable of sliding in a radial slot in the piece B^ as 
clearly shown in the sectional view. This piece B is sup- 
ported by the shaft D^ which turns in a bearing extending 
through the lower part of the large pin. R Z, drawn through 
the center of D^ is the line of motion of the tool slide. The 



APPLIED MECHANICS. 



835 



connecting-rod, actuating the tool slide, is pivoted to the 
stud 6 , which is clamped to piece B. The parts are lettered 




Fig. 3»3l. 

as in the two previous cuts, and the student should be able 
to study out the working of the mechanism without further 
explanation. 

1 457. The Adjustment of tlie Stroke. — The radial 
slot 7", in Fig. 361, to which the connecting-rod is attached, 
provides for the adjustment of the length of the stroke, and 
if the point of attachment of the rod to the tool slide is also 
made adjustable, the position of the stroke, as well as its 
length, can be changed. Thus, in the case of a shaping 
machine, it is not only desirable to regulate the distance 
passed over by the tool, but to have the stroke extend exactly 
to a certain point. These two adjustments are often required 
in mechanism where reciprocating pieces are employed. 



OTHER LINKAGES. 

1458. The universal joint, shown in Fig. 362, is 
used to connect two shafts, the center lines of which are in 
the same plane, but make an angle with one another. It is 
generally constructed in the following manner : Forks F^ F 
are fastened to the ends of the shafts A and B^ and have 
burrs ^, a tapped out to receive the studs 5, 5. The ends of 
these studs are turned cylindrical, and are a working fit, in 



836 



APPLIED MECHANICS. 



corresponding bearings in the ring R. The details of con- 
struction may be seen in the right-hand part of this figure. 




Fig. 862. 



In heavy machinery the forks are forged and welded to the 
shafts. 

1459. An objection to the universal joint is that the 
motion transmitted is not uniform. During one revolution 
the speed of the driven shaft varies twice between a velocity 
that is greater, and one that is less than the velocity of the 
driving shaft, and between these points there are four posi- 
tions where the velocity of the two shafts is the same. 
Suppose the driving shaft to revolve uniformly; the /easf 
speed of the driven shaft will be equal to the speed of the 
driving shaft, multiplied by the cosine of the angle between 
the two axes produced (the angle x in Fig. 362). Th.^ greatest 
speed will be equal to the speed of the driver, multiplied by 

• . Thus, if the shafts revolve at the rate of 100 revo- 

cos X 

lutions per minute, and the angle between them is 30°, the 

least speed of the driven shaft will momentarily be at the rate 

of 100 X cos 30°= 100 X .86603 = 86. 6 revolutions per minute, 

and the greatest speed, — -3 = -— — — - = 115.47 revolutions 

cos oU .oDoUo 

per minute. 

1460. To obviate this trouble, which brings excessive 
wear and stresses on the working parts, the double univer- 
sal joint is used, as shown in Fig. 363. Let A and B be 
the two shafts to be connected. Draw their center lines, 
intersecting at 0^ and bisect the angle A dhy the line a 0, 




APPLIED MECHANICS. 837 

The center line e f oi the connecting shaft D must now be 
drawn perpendicular to a o. Care must be taken that the 
forks on the interme- 
diate shaft lie in the 
same plane. Thus 
constructed, a uni- 
form motion of A will 
give a varying mo- 
tion to D, which in 
turn will transmit to 
B the same motion 
as that of A. This "e" 

arrangement is often ^fig. 363. 

employed to connect parallel shafts, as would be the case in 
the figure if o d took the direction s t. In such a case, it 
makes no difference what angle is made by e y, except 
that, if the joints are expected to wear well, it should not 
be too great. 

1461. Watt's Parallel Motion. — A parallel mo- 
tion, more properly called a straight-line motion, is a 

link mechanism designed to guide a reciprocating piece, as 
a piston rod, in a straight line. In the early days of the 
steam engine, parallel motions were extensively used to guide 
the pump and piston rods, but are now seldom met with, 
except on steam-engine indicators, where they are employed 
to give a straight-line motion to the pencil. Very few par- 
allel motions produce an absolutely straight line, and it is 
customary to design them so that the middle and two 
extreme positions of the guided point will be in line. The 
best known motion is the one shown in Fig. 364, which was 
invented by James Watt, in 1784. The links A B and CD 
vibrate on their fixed centers A and D. The other ends, B 
and (7, are connected by the link C B, which has the point 
O so chosen that it will pass through three points, O^, O^ 
and (9^, in the straight line 5 S perpendicular to the links 
C D and A B when in their middle positions. When the 
point O is at the upper extremity of its motion at O^^ the 



838 



APPLIED MECHANICS. 



linkage assumes the position A B^C^D\ at the lower extrem- 
ity it assumes the position A B^ C^ D. 

1462. Having given the length O^ O^ of the stroke, 
and (9, the middle position of the guided point, the center 
of one lever as Ay and the perpendicular distance between 



^ A 




8 

Fig. 364. 

the levers when in mid position, the motion may be laid out 
as follows: Let 5 5 be the path of the guided point, O its 
middle position, A the given center, and A B and CD indefi- 
nite parallel lines, representing the middle positions of the 
levers. From in^ where A B intersects wS vS, lay off upon 
5 5 the distance in a^ equal to \ of the stroke. Join A with 
a and draw an indefinite line a /?, perpendicular to A a. The 
point By where a B intersects line A B, is the right-hand 
extremity of lever A B, and the lower extremity of link C B. 
The point (7 is obtained by drawing an indefinite line through 
B and 0\ where it intersects the line CD will be the point. 



i 



APPLIED MECHANICS. 839 

To find center D lay off n b ^^\ stroke ; connect C and b^ 
and from b draw an indefinite line perpendicular \.o C b\ the 
center will be at its intersection with C D. 

If the positions of both centers should be known, mark 
points a and b as before. Draw A a and b D^ and through 
b and a draw perpendiculars to these lines ; the points B and 
C^ where they intersect the center lines of the levers, are the 
extremities of these levers. Join B and C by the link B C, 
and the point O, where the center line of this link cuts the 
line of motion 5 5, is the position of the guided point O on 
the link B C, 

CAMS. 

1463. A cam is a turning or sliding piece, which, by 
the shape of its curved edge, or a groove in its surface, 
imparts a variable or intermittent motion to a roller, lever, 
rod, or other moving part. 

1464. General Case. — Fig. 365 represents the general 
case for a plate cam. The cam C is supposed to turn in a 
right-handed direction about the axis B, and to transmit a 
variable motion through the roller A^ and the lever Z, to the 
rod R. The lever swings on the axis 0^, the roller moving 
up or down on the arc to 6, as the cam revolves. The 
roller is held in contact with the cam by its own weight and 
that of the lever and the rod. 

Suppose the location of the cam shaft B and the rod R 
to be known, and that the cam is to revolve uniformly right- 
handed and impart motion to the rod, so that during the 
first half of every revolution it will move uniformly down- 
wards, during the next quarter turn it will remain stationary, 
and during the last quarter it will return to its former posi- 
tion with a uniform motion. 

1465. In order to determine the outline of the cam 
upon which the circumference of the roller bears, it is neces- 
sary to find an outline which will give the center of the roller 
the required motion. Then, by placing the point of the 



840 



APPLIED MECHANICS. 



compasses at different places on this outline, and striking 
arcs inside of it, with radii equal to the radius of the roller, 
the curve for the actual cam can be drawn, the curves being 
tangent to the arcs, and parallel to the first outline. 

In this case, the roller A moves in an arc directly over 
the center of B. Knowing the distance the rod R is to 
move, we must so choose the point 0^ and the throw of the 
cam, that is, the distance that A is to move, that (see figure) 
the movement of R will be to the movement of ^ as r is to a. 

Now, with 0^ as a center and a radius equal to a, describe 




Fig. 365. 

the arc O-S-6, in which the center of the roller is to move, 
and mark the highest and lowest points and 6 of the roller. 
The lower point should not be near enough the shaft to 
allow the roller to strike the hub of the cam. 

1466. It evidently makes no difference with the rela- 
tive motions of the cam and roller whether the cam turns 



APPLIED MECHANICS. 841 

right-handed, and the lever remains with its axis at 0^^ or 
whether the cam is assumed to be stationary, and the lever 
and roller move left-handed in a circle about the center B. 
This latter process will be adopted. 

With B as the center, draw a circle through 0^ and space 
it into a number of equal parts, say 12, and number the 
divisions around to the left. Now, assume the lever to move 
around the axis ^ in a left-handed direction. It will take 
positions 0^-1, 0^-2, 0^-3, etc. Hence, using these several 
points on the outer circle as centers, and with radii equal to 
a, the length of the lever, describe a series of arcs corres- 
ponding to the original arc 0-3-6. Number these arcs 1, 2, 3^ 
etc., to correspond with the numbers on the outer circle. 

During the first half -turn of the cam, or, what is the same 
thing, while the lever is moving from its position at 0^ to 0^ 
on the outer circle, the center of the roller must move uni- 
formly from its outer to its inner position. Hence, draw 
the chord of the original arc 0-3-6, and divide it into six 
equal parts, numbering them towards the center as shown. 
Then, with ^ as a center, describe an arc through point i, 
intersecting arc No. 1 in point 1^. Now, sweep arcs through 
the other points, getting ^^, 3^, Jf^, etc., which are all points 
in the curve of the cam outline, for the center of the roller. 
From 0^ to 0^ (on the original circle), the center of the roller 
remains at a constant distance from B; hence, 6 and 9 must 
be connected by a circular arc. From 0^ to 0^^ the points 
are found as before by dividing the chord 0-6 into three 
equal parts and numbering them as shown, the numbers 
running outwards. 

The final steps are to draw the cam outline for the center 
of the roller through points 1^, 2^, ^,, ^j, etc. ; then, draw the 
outline for the cam itself parallel to it, as explained at first. 
This is easily done by setting the dividers to describe a 
circle whose radius shall be the same as that of the roller A. 

Then, with various points on the curve 0-1^-2^-3^ 11^ as 

centers (the more, the better), describe short arcs as shown. 
By aid of the irregular curve, draw a curve which shall be 
tangent to the series of short arcs; it will be the required 



842 APPLIED MECHANICS. 

outline of the cam, and will be parallel to the curve 

(?.i,-a.-^, — 11,. 

1467. The question sometimes arises in designing cams 
of this nature, whether it is the chord 0-6 or the arc 0-6 
that should be divided to give the roller the proper outward 
and inward motions. For all practical purposes either way 
is sufficiently exact, but neither is quite correct, though it is 
better to space the chord. The exact way would be to draw 
the rod R and the roller in the different positions desired, 
and then design the cam to meet the roller at these 
points. 

1468. The cam shown in Fig. 366 differs in principle 
from the preceding one only in that the roller moves in a 
straight line, passing to one side of the center B of the shaft. 
Let it be required to design a cam of this nature to revolve 
right-handed, and which shall cause the roller A and rod R 
to rise with a uniform motion to a distance h during two- 
thirds of a revolution. When the roller reaches its highest 
point, it is to drop at once to its original position, and to 
remain there during the remainder of the revolution. As- 
sume the distance from the center B to the center line of R 
to be equal to r. 

With r as a radius, describe a circle about B^ as shown. 
The center line of the rod will be tangent to this circle in 
all positions. With the same center and a radius equal 
to B A^, A^ being the extreme outward position of the 
roller, describe the outside circle A^ j^-8 A^. Divide this 
circle into some convenient number of equal parts, the num- 
ber depending upon the fraction of a revolution required for 
the different periods of motion. Since the roller is to rise 
during two-thirds of a revolution, we may use 12 divisions 
as before, thus giving f x 13 = 8 whole divisions for the first 
period. 

Now, proceeding as before, by assuming the rod to move 
about the cam to the left, its positions when at the points of 
division A^ A^^ etc., will be represented by drawing lines 



APPLIED MECHANICS. 



843 



through these points, and tangent to the inner circle, whose 
radius is r. A and A^ are the two extreme positions of the 
roller. Divide the line A A^ into eight equal parts, num- 
bering them from the inside outwards, since the first move- 
ment of the roller is outwards. With ^5 as a center, draw 
concentric arcs through these points, intersecting the tan- 
gents at ij, ^,, <^j, etc. At point 8 on the outer circle, the 
roller drops along the line 8-8^^ the point 8^ being deter- 
mined by drawing an arc about B with a radius D A. From 




////X/m 
1 1 p. / ,//. 



5V-\-V-\-V-' 



point 8 back to A the rod is at rest. The true cam outline is 
now to be found, as was done in the last example, by striking 
small arcs from points on the curve A i,-^,-^,->^,-5,-6,-7,-<^-<5, 



844 



APPLIED MECHANICS. 



as centers. This cam can revolve in only one direction 
when operating the rod R. 



1469. Harmonic-Motion Cams. — If a cam is re- 
quired to give a rapid motion between two points, without 
regard to the khzd of motion, its surface should be laid out 
so as to gradually accelerate the roller at the start, and to 
gradually retard it at the end of its motion, in order that 
the movement may be as smooth and free from shocks as 
possible. For this reason cams are frequently designed to 
produce harmonic motion — that is, a uniform motion of the 
cam produces a motion of the roller like that of the slotted 
cross-head in Fig. 350. The diagram in Fig. 367 shows how 
this latter motion is plotted. Let A ChQ the stroke line of 




the cross-head, and ABC the crank-pin circle, which is 
divided into a number of equal spaces by the points a, b, c, d, 
etc. Dropping perpendiculars from these points, we ob- 
tain points i, ^, S^ ^ etc., on the stroke line. The spaces 
between the latter points represent the distances traversed 
by the cross-head while the crank-pin moves through the 
equal spaces a 0, b c^ c d^ etc. It will be seen that the dis- 
tances increase from points 1 to 5, and decrease from 5 to 9. 

1470. To apply the motion to the two cams previously 
taken up, we should simply have to lay off the distances 1-2^ 
2-S, S-i, etc., on the chord 0-6, in Fig. 365, or the line A^A, 
in Fig. 366, in place of the equal spaces used in these figures. 



APPLIED MECHANICS. 



845 



1 a 




Fig. 368, which represents the left side of the first cam con- 
sidered, laid out in this 
way, shows a convenient 
method of spacing. Upon 
the chord 0-6, as a diame- 
ter, draw the semi-circle 
d 6; divide it into a suit- 
able number of equal parts 
and project these divisions 
by straight lines to the / 
chord 0-6. Through the 
points of intersection i, ^, j^ 
S, 4) etc. , and with ^ as a * 
center, describe the arcs 
1-1^, 2-2^, S-3^, etc., and 
complete the cam outline, 
as before. Fig. 368. 

1471. Positive-Motion Cams. — The cams thus far 
considered can drive the roller in one direction only, making 
a spring or weight necessary to keep the two in contact. 
If the cam plate should extend beyond the roller, however, 
and a groove should be cut in it for the roller to run in, the 
motion of the roller would be positive in both directions. 

1472. The word positive, when applied to a mechan- 
ism, has a different meaning from any heretofore given to 
it. A mechanism so constructed that nothing short of 
actual breakage of some one of its parts can keep it from 
working properly when motion is imparted to one of the 
links which operates it is called a positive mectiaiiisiii, 
or a positive gear, when speaking of valve gears, and the 
motion produced is called a positive motion. Those 
mechanisms which depend for their operation upon the rais- 
ing or lowering of a weight (i. e., upon gravity), or the 
action of a spring, are termed non-positive or force- 
closed meclianisms. Non-positive mechanisms, although 
extensively used, are of a lower order of mechanical excel- 
lence than positive mechanisms, and, other things being 



846 



APPLIED MECHANICS. 



equal, a positive motion should be chosen when designing 
a mechanism, since a non-positive one will refuse to work if 
the weight or the part operated by the spring should get 
** caught." 

1473. Sometimes a positive motion is secured with a 
plate cam by causing it to revolve between two rollers 
rigidly connected, as illustrated in Fig. 369. The rollers A 
and E turn upon pins in one end of the rod R. The rod 
is slotted between the rollers, so that the cam shaft may 




pass through it and still allow the rod to move to the right 
and left. The center line of motion of the rollers passes 
through the center B of the shaft, whatever the position of 
the cam or rod. 

1474. Let it be required that during one-quarter of a 
revolution of the cam the rod shall move to the right or 
left, according to the position of the cam at the start ; that 
during the next quarter the rod shall be held stationary; 
that during the third quarter it shall move to its original 
position, where it is to remain for the rest of the revolution. 

Draw the line D //, and mark the point B. Lay off a dis- 
tance B F, such that when the center of one of the rollers is 



APPLIED MECHANICS. 847 

at F the hub of the cam will not interfere with the roller. 
Lay off the distance F H beyond F^ equal to the required 
movement of the rod. With ^5 as a center and a radius 
equal to B H^ describe a circle. Divide the circle into four 
quadrants by the lines a b and c d, making angles of 45° with 
D H^ and with center B and radius B F strike an arc as 
shown. That part of the arc included in the lower quadrant, 
and that part of the outer circle included in the upper 
quadrant, form two parts of the required outline, and the 
distance between the centers of the rollers must be equal 
toZ^i^ 

Now, for the two side quadrants, it is evident that the dis- 
tance between any two diametrically opposite points on the 
outline must be equal. Thus, the distance from i' to 6\ 
from 2' to 4", from S' to S\ must all be equal to the distance 
D F. These points are obtained by laying off points on F H 
and drawing arcs through them, proceeding exactly as with 
the other cams ; but^ in order to have the curves correct^ point 
1 must be as far from F as 5 is from H; point 2 as far from 
F as Jf. is from H, etc. The harmonic motion curve fulfils 
this condition, and was used in this case. The other points 
of construction should be understood from what has gone 
before. 

1475. A third kind of positive-motion cam consists of 
a cylinder having a groove on the surface, which imparts 
motion to the roller in a plane parallel to the axis of 
the cam. 

Suppose that during one-half a revolution of the cylinder, 
in Fig. 370, the arm is to vibrate to the left and back once, 
as indicated, and that during the other half revolution it is 
not to move. Let the motion of the roller be harmonic. 

The problem consists in finding the center line of the 
groove, from which, by striking arcs, the sides against which 
the roller bears can be determined. To lay out the curves, 
assume the surface of the cylinder to be unwrapped, or de- 
veloped, as represented by the figure a b d c, which repre- 
sents only a little more than one-half of the length of the 



848 



APPLIED MECHANICS. 



surface, in order to save room. Draw a line J/ TV through 
the center of this strip, of a length equal to the circumfer- 
ence of the cylinder, and divide it into a number of equal 
parts, as indicated by O^ O^, (9, O^^ etc. Also, draw two 
lines, 6" T and L P, parallel to MN^ and at a distance from it 
equal to one-half the desired stroke of the roller. Now, with 




Fig. 370. 



a radius A (left-hand figure), and centers O^, (9„ (9,, 
etc., strike arcs 0^ i, ^, 3^ etc., and plot the curve by the 
aid of these arcs. The line O K L is the center line curve 
for one-half of the cam, the other half being like the first. 
The outline can easily be transferred to the cylinder itself. 



APPLIED MECHANICS. 



849 



BELTING. 

1476. Belts running over pulleys form a convenient 
means for transmitting power, but they are not suited to 
transmit a precise velocity ratio, owing to their tendency to 
stretch and slip on the pulleys. For driving machinery, 
however, this freedom to stretch and slip is an advantage, 
since it prevents shocks that are liable to occur when a 
machine is thrown suddenly into gear, or when there is a 
sudden fluctuation in the load. 

1 477. Velocity Ratio. — Let four pulleys be connected 
by belt, as shown in Fig. 371. Let D^ and D^ be the diam- 
eters of the drivers, F^ and F^ of the followers, N^ and N^ 




NiHev, 



Ut Bev, 



Ug Rev, 



Fig. 371. 

the number of revolutions per minute of the drivers, and n^ 
and n^ of the followers. The two middle pulleys are keyed 
to the same shaft and revolve together. 

Consider first the pulleys whose diameters are D^ and F^, 
Assuming that there is no slipping or stretching of the belt, 
the circumferential speeds of the pulleys will be the same as 
the velocity of the belts passing over them. Hence, D^ X 
3.1416 X -A^i, the circumferential speed of the first driver, 
= ^, X 3.1416 X n^y the circumferential speed of the first 
follower. Canceling 3.1416 from both sides of the equatior., 
we have D^ N^ = F^ n^^ or, dividing by D^ and ?2„ 

N F 

^ = -^. (136.) 

That is, the speeds or numbers of revolutions of two con 
nected pulleys are inversely proportional to their diameters. 

D. 0. III.— 12 



850 APPLIED MECHANICS. 

147'8. A short way of applying this principle is by 
the following rule for two pulleys: 

Rule. — Multiply together the number of revolutions and 
diameter of one pulley^ and divide by the given number of 
revolutions^ or given diameter^ of the other pulley. The result 
will be the required diameter or number of revolutions. 

Example. — A pulley 30 inches in diameter, making 210 revolutions 
per minute, drives a second pulley 14 inches in diameter. How many 
revolutions per minute does the latter pulley make ? 

Solution.— 30 X 210 == 6,300, and 6,300 -j- 14 = 450 revolutions. Ans. 

Example. — The driving pulley of a machine is one foot in diameter 
and must make 750 revolutions in 5 minutes. What size pulley should 
be used on the driving shaft, if its speed is 143 revolutions per minute ? 

Solution. — In all examples of this kind the speeds and diameters 
must be reduced to the same units. 750 rev. in 5 min. = 750 -4- 5 = 150 
rev. per min. ; one foot = 12 in. Hence, 12 X 150 = 1,800, and 1,800 -r- 
143 = 12.6 in., nearly. Ans. 

1479. From the equation D^N^ = F^n^^ derived above, 

D N 
we obtain, by dividing by F^^ n^ = — W~^' ^"^ ^^^ manner, 

taking the other two pulleys in Fig. 371, we obtain N^ = 

F n 
' ^ . But the two middle pulleys revolve together, so 

that the values of N^ and n^ are equal, and may be placed 

D N F n 
equal to each other ; thus, — ^^ = — \^t or, multiplying 

by F^ and A, N^ D^ D, = n^ F^ F^. (1 37.) 

1 480. That is, the speed of the first pulley^ multiplied 
by the diameter of each of the drivers^ equals the speed of the 
last pulley^ multiplied by the diameter of each follower. 

This formula is in most cases convenient to apply as it 

stands. 

Example. — Referring to Fig. 371, let the diameters of the drivers be 
32 in., the diameter of the first follower be one foot, and of the second 
follower 15 inches. If the first shaft has a speed of 60 revolutions per 
minute, what is the speed of the last shaft ? 

Solution.— Substituting in formula 137, 60 X 32 x 32 = ;?a X 12 X 

• J. ^ ^o .^ 60x32x32 ^,,, 

15, or Qividmg by 12 X 15. n-i = — — = 341^ rev. per mm. Ans 



APPLIED MECHANICS. 851 

Example. — An emery grinder is to be set up to run at 1,200 revolu- 
tions per minute. The countershaft (corresponding to the middle 
shaft in Fig. 371) has pulleys 20 and 8 inches in diameter. If the pulley 
on the grinder is 6 inches in diameter, what size pulley must be used 
on the main line shaft, its speed being 180 revolutions per minute ? 

Solution.— Substituting in formula 137, 180 X -C>i X 20 = 1,200 X 8 

X 6, or Dx = ' ^^Q 2Q — = 16 inches. Ans. 

1481. When the speeds of the first and last shafts are 
given, and the diameters of all the pulleys are to be found, 
the following method is convenient : 

Rule. — Divide the higher speed by the lower. If two 
pulleys are to be iiscd^ this will be tJie ratio of their diameters. 

If four pulleys are required^ find two numbers whose product 
equals the above quotient. One of these numbers will be the 
ratio of the diameters of one pair of pulleys^ and the other 
number of tJie other pair. 

Example. — It is required to run a machine 1,G00 revolutions per 
minute, the driving shaft making 320 revolutions per minute. What 
size pulleys are required, {a) when two pulleys are used; {p) when four 
pulleys are used ? 

Solution. — {a) 1,600 -> 320 = 5. The two pulleys must, therefore, 
be in the ratio of 5 to 1, the driving pulley being 5 tim.es as large as 
the driven pulley, since the latter has the greater speed. We will 
assume diameters of 30 and 6 inches. 

{J)) 2| X 2 = 5. One pair of pulleys must be in the ratio of 2^ to 1, 
and the other pair of 2 to 1. . We will assume diameters of 25 and 
10 inches for one pair, and of 12 and 6 inches for the other pair. 

1482. Direction of Rotation. — It will.be noticed, by 
referring to Fig. 371, that the pulleys are connected by 
open belts where indicated by full lines, and by crossed 

belts where indicated by dotted lines. Pulleys connected by 
open belts turn in the same direction^ and those connected by 
crossed belts hi opposite directions. 



EXAMPLES FOR PRACTICE. 

1. A driving pulley is 54 inches in diameter, and a driven pulley 
which runs at 112 revolutions per minute is 2| feet in diameter. What 
is the speed of the driving shaft ? Ans. 62.22 rev. per min. 

2. The fly-wheel of an engme running at 180 revolutions per minute 
is 8 ft. 5 in. in diameter. What should be the diameter of the pulley 



852 



APPLIED MECHANICS. 



which it drives, if the required speed of the latter is 600 revolutions 
per minute ? Ans. 30^^^ in. , nearly. 

3. In Fig. 371, let the diameters of the two drivers be 48 and 25 
inches, and of the two followers 16 and 12 inches. If the last pulley's 
shaft rotates 800 times in 5 minutes, what is the speed per minute of 
the first shaft ? Ans. 25.6 rev. per min. 

4. If the first pulley in Fig. 371 turns right-handed, and is con- 
nected with the second by a crossed belt, and the third with the fourth 
by an open belt, in what direction would the last pulley turn ? In 
what direction would it turn if two crossed belts were used ? 

5. A machine is to be belted through a countershaft, so as to run 
at 1,200 revolutions per minute, the speed of the driving shaft being 
120 revolutions per minute. Find three ratios that could be used for 
each pair of pulleys. ^5 : 1 and 2 : 1. 

Ans. 3 4 : 1 and 2i : 1. 
( 3^ : 1 and 3 : 1. 

6. An emery grinder is to be set to run at 1,400 revolutions per 
minute. The countershaft has pulleys 30 and 8 inches in diameter. 
The pulley on the grinder is 7 inches in diameter. What size pulley 
should be used on the main line shaft, its speed being 185 revolutions 
per minute ? Ans. 14^ inches. 



POWER TRANSMISSION BY BELT. 

1483. The Effective Pull.— In Fig. 372, let D and 

F be two pulleys connected by a belt, D being the driver 
and F the follov»^er. To avoid undue slipping, the belt must 
^Tension =Tz*^- be drawn tight. This will pro- 
duce a tension in the upper 
and lower parts which we call 
T^ and 7",, respectively. 

Suppose the two pulleys to 
be stationary and that the 
belt is put on with a certain 
tension. Then, 7"^ will equal T^. But if the pulley i? should 
be turned in the direction of the curved arrow it would tend 
to stretch the lower part of the belt, increasing its tension 
still more, while the tension of the upper part would be 
diminished an equal amount. This would go on until the 
difference of the tensions was sufficient to start pulley F, 

This difference (7"^ — T^ is the pull that does the work in 
transmitting power, and is called the effective pull. In any 




Fig. 372. 




APPLIED MECHANICS. 



853 



case, therefore, the number of foot-pounds of work trans- 
mitted by a belt must equal the effective pull of the belt in 
pounds times the number of feet passed through ; or, tak- 
ing a minute as the unit of time, tJie power transmitted in 
foot-pounds per minute = the effective pull X the velocity in 
feet per minute, 

1484. To show clearly how the effective pull enters into 

calculations of power transmissions, two examples will be 

solved. 

Example. — The diameter of the driving pulley D is 36 inches. It 
makes 150 revolutions per minute and carries a belt transmitting 
6 horsepower. What is the effective pull of the belt ? 

Solution. — Velocity of the belt in feet per minute = 
150x36x3.1416 



12 



= 1,413.72. 



This, multiplied by the effective pull in pounds must equal the foot- 
pounds of work done per minute, or 6 X 33,000. Hence, letting P = 
the effective piill, and equating these expressions, we have 
P X 1,413.72 = 6 X 33,000 = 198,000, 

„ 198,000 ..^nAiK 1 

or -P— -, iiQ ryo — 140.06 lb., nearly. 

i,4:l0. lu 

Example. — A pulley 4 feet in diameter is driven at 100 revolutions 
per minute, and transmits power to another pulley by means of a belt 
without slip. If the tension on the driving side of the belt is 400 
pounds, and on the slack side is 100 pounds, what is the horsepower 
transmitted ? 

Solution.—/' = ( Ti - Ta) = 400 — 100 = 300 lb. Horsepower trans- 
mitted X 33,000 = foot-pounds of work done per minute; or 
H.P. X 33,000 = 300 X 100 X 4 X 3.1416, whence 
300x100x4x3.1416 



H. P. = 



= 11.424. Ans. 



33,000 

1 485. To Determine the Widtli of Belt. — A belt 
should be wide enough so that it will bear safely and for a 
reasonable length of time the greatest tension that will be 
put upon it. This will be the tension T^ of the driving side 
of the belt. As belts are usually laced, or fastened with 
metallic fasteners, both of which require holes to be 
punched in the ends, it is customary to use the breaking 
strength through the lace holes, divided by a suitable factor 
of safety, as the greatest allowable tension. The average 
breaking strength for sing^le leather belts, through the lace 



854 



APPLIED MECHANICS. 



holes, is 200 pounds per inch of width. This divided by 
three, which is a suitable factor of safety for belting, gives 
GOf pounds. Thus, in the last example, the tension of the 
driving side of the belt was assumed to be 400 pounds. 
Hence, using GGf pounds as the safe working stress per inch 

of width, a belt —rxir =■ G inches wide would be required. 

This tension 7"^, for any particular case, depends upon 
three things — viz. , the effective pull of the belt, the coefficient 
of friction between the belt and pulley, and the size of the 
arc of contact of the belt on the smaller pulley. As the 
equations involving these quantities are somewhat compli- 
cated. Table 33 has been calculated. It will afford a con- 
venient means for finding not only the width of belt for a 
given horsepower, but the horsepower for a given width as 
well. In the first column, the arc covered by the belt is 
stated in degrees, and in the second column in fractional and 
decimal parts of the circumference covered. The third 
column gives the greatest allowable values of {T^ — /"J, or 
the effective pull, per inch of width, for single leather belts 
having any arc of contact. It was computed by assuming 
a value for T^ of QQ^ pounds, and a coefficient of friction of 
.27. This latter has been found by experiment to be a fair 
value to use for leather belts running over cast iron pulleys, 
under conditions met with in practice. 

TABLE 33. 



Arc Covered by Belt. 


Allowable Value 
of Effective Pull, 


Degrees. 


Fraction of 
Circumference. 


or ( Ti — Ti) per 
Inch of Width. 


90 


i=.25 


23.0 


112i 


-iV=-312 


27.4 


120 


|-=.333 


28.8 


135 


f ==.375 


31.3 


150 


A =-417 


33.8 


157i 


tV=.437 


34.9 


180) 
or over. ) 


i=.50 


38.1 



APPLIED MECHANICS. 855 

1486. To use the table in finding the width of a sin- 
gle leather belt required for transmitting a given horse- 
power, we have the following rule : 

Rule. — Compute the effective pull of the belt. Divide the 
result by the suitable effective pull per inch of width^ as given 
in Table 33; the quotient will be the width of belt required^ in 
inches. 

Example. — What width of single belt is needed to transmit 20 
horsepower with contact on the small pulley of f of the circumference 
and a speed of 1,500 feet per minute ? 

Solution.— First finding the effective pull, P X 1,500 = 30 X 33.000, 

20 X 33,000 _ 

1,500 ~ 

440 
Hence, 440 -*- ( 7\ — Tlj), from the table, = „ . „ = 14 inches, nearly. 

ol. o 

A 14-inch belt would be used. Ans. 

1487. To Determine ttie Horsepo-wer tliat a 
Belt Will Transmit. — The process for a single belt must 
evidently be just the reverse of the preceding. It is as 
follows : 

Rule. — Multiply together a suitable value for the 
effective pull, taken from Table 33, the width of the belt 
in inches, and its velocity in feet per minute. The result 
divided by 33,000 will be the horsepower that the belt will 
transmit. 

Example. — What horsepower will a one-inch belt transmit with a 
speed of 900 feet per minute and an arc of contact of 180° ? 

Solution.— Ti - T^, from the table, = 38.1. 38.1 x 1 X 900 = 34,290, 
which, divided by 33,000, gives 1.04 horsepower, nearly. Ans. 

1488. General Rule for Belting. — From the last 
example, we see that a single belt traveling 900 feet 
a minute will transmit one horsepower per inch of 
width when the arc of contact on the smaller pulley 
does not vary much from 180°. This is used by many 
engineers as a general rule for belting, to be applied to 
all cases. 



856 APPLIED MECHANICS. 

The following three formulas express the operations that 
would be performed in applying this rule : 

Let H ^ horsepower to be transmitted; 
W ^= width of belt in inches; 
»S = belt speed in feet per minute. 

Then, ^=S- (*3®-> 

W=^. (139.) 

5 = -^. (140.) 

Example. — Two pulleys, 48 inches in diameter, are to be connected 
by a single belt, and make 200 revolutions per minute. If 40 horse- 
power is to be transmitted, what must be the width of belt ? 

o o^t, t. 14. ^ 200X48X3.1416 « kio ^ ^ 

Solution. — The belt speed = ^75 = 2,513 feet per 

1« 

900 X 40 
minute, about. Applying formula 139, W= ^ -■.„ =14.3 inches. 

Ans. 

A 14-inch belt might safely be used, since the rule gives a liberal 
width when the pulleys are of equal size. 

Example. — ^What size pulleys should be used for a 4-inch belt, which 
is to connect two shafts running at 400 revolutions per minute and 
transmit 14 horsepower ? Both pulleys are of the same size. 

900 X 14 
Solution. — By formula 140, 5= ^ = 3,150 feet per minute. 

Since this speed = the circumference of the required pulley in feet 

X 400, we have circumference of pulley = —^ — -^^ = 94.5 inches; 

94 5 
diameter = „ . ' ._ = 30 inches. Ans. 
0.14I0 

1489. Double Belts. — Double belts are made of two 
single belts cemented and riveted together their whole 
length, and are used where much power is to be transmitted. 
As the formulas for single belts are based upon the strength 
through the rivet holes, a double belt, which is twice as thick, 
should be able to transmit twice as much power as a single 
belt, and, in fact, more than this, where, as is quite com- 
mon, the ends of the belt are glued instead of being 
laced. 



APPLIED MECHANICS. 857 

Where double belts are used upon small pulleys, however, 

the contact with the pulley face is less perfect than it would 

be if a single belt were used, owing to the greater rigidity 

of the former. More work is also required to bend the belt 

as it runs over the pulley than in the case of the thinner and 

more pliable belt, and the centrifugal force tending to throw 

the belt from the pulley also increases with the thickness. 

Moreover, in practice, it is seldom that a double belt is put 

on with twice the tension of a single belt. For these 

reasons, the width of a double belt required to transmit a 

given horsepower is generally assumed to be seven-tenths 

the width of a single belt to transmit the same power. 

Upon this basis, formulas 138, 139, and 140 become, for 

double belts, 

WS 
H = ^. (141.) 



630* 
W 



W=^-^. (142.) 



^ 630/f ,^ ._ . 



EXAMPLES FOR PRACTICE. 

1. If the effective pull on a belt per inch of width is 50 pounds, and 
the belt passes over a pulley 36 inches in diameter, which makes 160 
revolutions per minute, how wide should the belt be to transmit 12 
horsepower ? Ans. 5^ inches. 

2. What width of single belting should be used to transmit 5 horse- 
power, when the belt speed is 2,000 feet per minute, and the arc of con- 
tact on the smaller pulley is 90° ? Ans. 3^ + inches. 

3. Using the general rule, find the horsepower that a 16-inch single 
belt will transmit, the belt speed being 1,000 feet per minute. 

Ans. 17.8 H. P. 

4. Calculating from Table 33, how much power could the above 
belt be depended upon to transmit if the arc of contact on the smaller 
pulley is -J- of the circumference ? Ans. 14 H. P., nearly. 

5. Required the diameter of pulleys necessary to enable a 10-inch 
belt to transmit 9 horsepower at 125 revolutions per minute, both 
pulleys being of the same size. Ans. 2 ft., nearly. 

6. How much power would the belt in example 3 transmit, if the 
belt were double ? Ans. 25.4 H. P., nearly. 



858 



APPLIED MECHANICS. 



SPEED CONES. 

1 490. Speed cones are used for varying the velocity of 
a shaft or other rotating piece driven by a belt. Their 
method of operation will be clearly seen in Figs. 373 to 376. 




LrV 



Fig. 3:3. 



Fig. 374. 



Figs. 373 and 374 show continuous cones and conoids, re- 
spectively, the former being suitable for crossed, and the 
latter for open belts, where the speed of the driver shaft 
can be raised gradually by shifting the belt. Figs. 375 and 
376 show sets of stepped pulleys. As flat belts tend to climb 




Fig. 375. 



Fig. 376. 



a conical pulley, continuous cones or conoids require 
special provision for keeping the belt in place. For this 
reason the stepped cones are generally used. Whenever 



i 



APPLIED MECHANICS. 



85li 



iV Revolutions. 
A 



Y(i 



possible, It is desirable to have both pulleys alike, so that 
they can be cast from one pattern, and in what follows it will 
be assumed that this is to be done. 

1491. Cotitiniious Cones for Crossed Belts. — Let 

A and B, in Fig. 377, re- 
present two speed cones of 
the same size, having the 
diameters of the large and 
small ends equal to D and 
d^ respectively. A is the 
driver, and revolves at a 
constant number of revolu- 
tions N^ while the speed of 
B is n^ or n^, according as 
the belt runs at the small 
or large end. 

It is assumed that pul- 
ley B is to have a range of 
speeds between n^ and ii^ revolutions, n^ being the greater. 

Since A and B are to be of the same size, it will be found 
that a certain relation must exist between N and n^ and n^ 
as follows : 




Fig. 377. 



From formula 136, 



ND 



n^d^ or D = 



and Nd= n^D, or D 



n^d 

_Nd 



(144.) 
(145.) 



Equating 144 and 145, 



n, d Nd 



N 



n. 



or 



(146.) 

That is, N must equal the square root of the product of n^ 
and n^. 

1492. Having determined TV, the relation between E 
arid ^can be found. From 144 and 145 we have: 

iV^= -^ and A^= -^. 



860 

Equating, 
Hence, 



APPLIED MECHANICS. 



= -^;t-, or n^ d^ = n„ D \ 



D 



d 



D 



dV^K 



(147.) 



That is^ the large diameter must equal the small diameter ^ 
multiplied by the square root of the qiwiient of n^ divided 
by n^. 



1493. Speed cones, to be properly designed, should 
have their diameters at different points, so proportioned that 
the belt will always have the same length, when tightly 
drawn, whatever its position. Let d and D' represent the 
diameters of two pulleys, connected by a crossed belt, whose 

centers are O and C^ re- 
spectively. (See Fig. 
378.) The length of the 
belt is 2 ^ ^+the length 
of the arcs from i^ to ^ 
and B to E. Tliese arcs 
Fig, 378. subtend equal angles, 

and supposing each to contain x degrees, the length of the belt 

= 2^^+ 3.1416 (^+i)')^|^, there being 360 degrees in a 

circle. Now, draw a line G H parallel to A B, and, about 
O and C as centers, describe dotted circles whose diameters 
are d' and D tangent to G H, representing two other pulleys. 
Then, for the length of the belt, we have % G H {=% A B) -\- 

X 




3.1416 {d' \-D)- 



360 



But (d' -f D) = {d+ D'), since what 



was taken from D' to make D was added to d to make d'. 
Hence, the length of the belt in each case is the same, and 
we have the rule that for crossed belts the sum of the corre- 
sponding diameters of two speed cones should be the same at 
all points. 

From this it follows that if two cones not exactly alike are 
to be driven by cross belt, it is only necessary to see that the 
sum of the diameters remains the same. 



APPLIED MECHANICS. 861 

Example. — Two continuous speed cones are to be designed to give 
a range of speed between 100 and 700 revolutions per minute. They 
are both to be alike in all respects. What must be the speed of the 
driving shaft, and the large diameter of the cones, assuming the small 
diameter to be 4 inches ? 

Solution.— From formula 146, 7V= ^/^T^za = |/100 X 700 = 
|/70,000 = 264.57 + revolutions per minute. Ans. 

From formula 147, D = d |/^ = 4i/!55 = 4 X 2.645 = 10.58 
inches. Ans. ^"^ 

1494. When an open belt is used, the values of TV, 
/?, and d are calculated as above, but for the other diameters 
a different rule is required. In Fig. 379, which is similar to 
Fig. 378, the pulleys are connected by an open belt. The 
figure is drawn so that D -{- d=z H ^ d\ the circles U and 
d' being made equal 
to represent the mid- 
dle sections of two 
cones. It is evident 
that the belt over D 
and d is longer than 
the one over U and 
d' , Hence, we see fig. 879. 

that the middle sections of speed cones for open belts must be 
larger proportionately than for crossed belts. This is indica- 
ted by jK in Fig. 374. Calling this middle diameter M^ and let- 
ting {7 be the distance in inches between the centers of the two 
shafts, upon which the cones are placed, it can be shown that 

^^^+-o8P-^r (148.) 

As the proof is a long one it will not be given. Having 
thus determined the end and middle diameters, the curve 
of the conoid may be taken as the arc of a circle passing 
through the extremities of the three diameters. 

Example. — What should be the middle diameter for the speed cone 
of the last example, having end diameters of 4 and 10.58 inches, when 
an open belt is to be used ? The distance between centers is 50 inches. 
Solution. — From formula 148, 

10.58 + 4 .08(10.58-4)'^ 
2 "^ 50 

= 7.29 + .07 = 7.36 inches. Ans. 




862 



APPLIED MECHANICS. 



E 



C 



D 



F 



1495. Stepped Pulleys. — When stepped pulleys, or 
cone pulleys, as they are more commonly called, are to be 

used, continuous cones should 
first be laid out as described, like 
A BCD, in Fig. 380. Then draw 
as many diameters, at equal dis- 
tances apart, as there are to be 
steps in the cone plus 1, as A Z>, 
£F, etc. These will serve as center 
lines for the different steps, which 
^^^- ^^' are to be drawn through the inter- 

sections of the above diameters, with the outside lines A B 
and D C, in the manner indicated. 

1496. It should be noted that, when speaking of a cone 
pulley as having a certain number of steps, the number of 
steps is one less than the number of pulleys on the cone. Thus, 
the cone, in Fig. 380, is a 4-step cone, and has five pulleys. 
Consequently, if a cone pulley (or cone) is spoken of as hav- 
ing fiv J steps, there are six pulleys and six changes of speed. 

If the distance between the axes of the pulleys to be con- 
nected by open belt be great, or if, as is sometimes the case, 
one of the axes be adjustable (the proper tension on the belt 
being obtained by the weight of the pulley on the adjust- 
able axis), the diameters can be calculated as though the belt 
were crossed. Otherwise, when designed for open belts they 
should be laid out as before described. 



EXAMPLES FOR PRACTICE. 

1. Two continuous speed cones are required to give a range of 
speed between 100 and 600 revolutions per minute. Assuming the 
large diameters of the cones to be 14 inches, {a) what must be the small 
diameters, and {p) the speed of the driving shaft ? Both cones are to 
be alike. . 3 {a) 5. 71 inches. 

• ( {b) 244.95 rev. 

2. In the above example, if the speed of the driving shaft were 260 
revolutions per minute, and the slowest speed of the driven cone 140 
revolutions, {a) what would be the greatest speed of the driven cone ? 
{b) What would be the ratio of the large and small diameters of the 
cones ^ . j (^) 482.86 rev. 

^^' \ ib) 1.86 : 1. 



APPLIED MECHANICS. 8d3 

3. What should be the middle diameter for the speed cones of 
example 1, when an open belt is used, the distance between centers being 
30 inches? - Ans. 10.04 inches. 



THE CARE AND USE OF BELTING. 

1497. Belts most commonly used are of leather, single 
and double. Canvas belts covered with rubber are some- 
times used, especially in damp places, where the moisture 
would ruin the leather. 

Leather belts are generally ran with the hair, or grain, 
side next the pulley. This side is harder and more liable to 
crack than the flesh side. By running it on the inside the 
tendency is to cramp or compress it as it passes over the 
pulley, while, if it ran on the outside, the tendency would be 
for it to stretch and crack. Moreover, as the flesh side is 
the stronger side, the life of the belt will be longer if the 
wear comes upon the weaker or grain side. 

1498. The lower side of a belt should be the driving 
side, the slack side running from the top of the driving pulley. 
The sag of the belt will then cause it to encompass a greater 
length of the circumference of both pulleys, as illustrated by 
the dotted lines in Fig. 372. Long belts, running in any 
direction other than the vertical, work better than short 
ones, as their weight holds them more firmly to their work. 

It is bad practice to use rosin to prevent slipping. It 
gums the belt, causes it to crack, and prevents slipping for 
only a short time. If a belt properly cared for persists in 
slipping, a wider belt or larger pulleys should be used, the 
latter to increase the belt speed. Belts, to be kept soft and 
pliable, should be oiled with castor or neatsfoot oil. 
Mineral oils are not good for the purpose. 

1499. Tightening or guide pulleys, whenever used 
to increase the length of contact between the belt and pulley 
or to tighten a belt, should be placed on the slack side, if 
possible. Thus placed, the extra friction of the guide pulley 
bearings and the wear and tear of the belt that Avould result 
from the greater tension of the driving side are avoided. 



864 



APPLIED MECHANICS. 



1500. Guiding Belts.— When belts are to be shifted 
from one pulley to another, or must be guided to prevent 
running off the pulley, the fork, or other device used for 
guiding, should be close to the driven pulley, and so placed 
as to guide the advancing side of the belt. 

This principle is sometimes made use of where pulleys have 
flanges to keep the belt on the pulleys. Where constructed 
with straight flanges, as in Fig. 381, if the belt has any inclina- 
tion to run on one side, its tendency is to crowd up against 
the flange as shown at ^, a. When constructed as in Fig. 823, 





Q 



Fig. 381. Fig. 382. 

however, with the flanges grooved as at c^ c^ the advancing 
side of the belt will be guided at b, just as it reaches the 
pulley, by contact with the thick portion of the flange, and 
during the rest of the way will not touch the flange at all. 

1501. In Fig. 383, is shown the 
arrangement for a belt shipper. G is the 
driving pulley and T and L are tight and 
loose pulleys on the driven shaft CD. B 
is the shipper, and can be moved parallel 
with C D. The acting surface, or face, 
of G is made straight to allow the belt 
to shift readily, and the faces of T and 
L are crowned, so that the belt will not 
tend to run off. 

1 502. The Climbing: of Belts.— 

In Fig. 384, suppose the shafts ^, a to be 
parallel, and the pulley A to be cone- 
shaped. The right-hand side of the belt 



^ 



FTO 



\i> 



T L 

Fig. 383. 

will be pulled ahead more rapidly than the left-hand side, 



APPLIED MECHANICS. 



8G5 



because of the greater diameter and consequently greater 
speed of that part of the pulley. The belt will, therefore, 
leave its normal line at <f, and 
climb to the ''high" side of 
the pulley. This tendency is 
taken advantage of by crown- 
ing pulleys in the middle. 
Each side of the belt then tends 
to move towards the middle of 
the pulley; that is, the ten- 
dency is for the belt to stay on 
the pulle}/. 

Suppose, on the other hand, 
the shafts b^ b not to be paral- 
lel, the right-hand ends being 
nearer together. The belt will 
in that case pass spirally on 
the pulley B towards the 

**10W" side. Fig. 384. 




VIAWVVVVW 








1503. Belt Fastenings. — There are many good 
methods of fastening the ends of belts together, but lacing 
is generally used, as it is flexible like the belt itself, and runs 
noiselessly over the pulleys. The ends to be laced should be 

cut squarely across and the 
holes in each end for the lacings 
should be exactly opposite each 
other when the ends are brought 
together. Very narrow belts, 
or belts having only a small 
amount of power to transmit, 
usually have only one row of 
holes punched in each end, as 
in Fig. 385. A is the outside 

To 



B 



Fig. 385. 
of the belt, and B the side running next the pulley, 
lace, the lacing should be drawn half way through one of 
the middle holes, from the under side, as at 1. The upper 
end should then be passed through ^, under the belt and up 

D. 0. III.— 13 



866 



APPLIED MECHANICS. 



4 




mlkt^ 



vAA/VlAAWVV^ 



through 3, back again through 2 and 3^ through ^ and up 
through 5, where an incision is made in one side of the lacing, 
forming a barb that will prevent the end from pulling 
through. The other side of the belt is laced with the other 
end, it first passing up through Jf. Unless the belt is very- 
narrow, the lacing of both sides should be carried on at once. 

1504. Fig. 386 shows a method of lacing where double 
lace holes are used, B being the side to run next the pulley. 
The lacing for the left side is begun at i, and continues 

through 2, 3, ^, 5, 6, 7, ^, 5, etc. 

A 6-inch belt should have 

seven holes, four in the row 

nearest the end, and a 10-inch 

belt, nine holes. The edge of 

the holes should not be nearer 

than f of an inch from the 

sides; and the holes should not 

be nearer than J of an inch 

from the ends of the belt. The 

second row should be at least 

If inches from the end. 

Fig. 386. Another method is to begin 

the lacing at one side instead of in the middle. This method 

will give the rows of lacing on the under side of the belt the 

same thickness all the way across. 



H 



^ 



mm'^ 



BELTS TO CONNECT NON-PARALLEL SHAFTS — 

GUIDE PULLEYS. 

1505. It very frequently happens that one shaft must 
drive another at an angle with it. Sometimes this involves 
the use of guide pulleys, and occasionally guide pulleys 
must be used to connect parallel shafts, where the shafts are 
near together, or there is some obstruction in the way. In 
all these cases, however, we have the general principle that 
the point at which the center of the belt is delivered from each 
pulley must lie in the middle plane of the other pulley. 



APPLIED MECHANICS. 



867 



The middle plane of a pulley, it will be understood, is a plane 
through the center of the pulley, perpendicular to its axis. 

Unless the shafting is to turn backwards at times, it is 
immaterial in what direction a belt leaves a pulley; but it 
must always be delivered into the plane of the pulley towards 
which it is running. If it is necessary for a belt to run 
backwards as well as forwards, it must leave in the plane of 
the pulley, also. This principle applies to the guide pulleys 
as well as to the main pulleys. 

1506. Shafts at Right Angles. 

— One of the most frequent cases met 

with is that of two shafts at right 

angles, but not intersecting, and the 

common method of connecting them 

is by means of a quarter-turn belt, 

shown in Fig. 387. Here, D is the 

driver, revolving in the direction 

shown, and F is the follower. The 

point at which the belt is delivered 

from the pulley D lies in the middle 

plane of the pulley i% that is, in the 

line b b\ also, the point at which the 

belt is delivered from the pulley /^lies 

in the middle plane of the pulley/?, or 

in the line a a. Thus arranged, the 

pulleys cannot run backwards, because 

</, the point of delivery of i% is not fig. 387. 

in the middle plane a a^ and e is not in the middle plane b b. 

1507". The following simple method of locating the 
pulleys for a quarter-turn belt may be used in practice: 
Let D, Fig. 387, be the driving pulley, and i^ the follower or 
driven pulley, which drives a machine. Locate the pulley 
D and the machine so that the pulleys D and F will be as 
near the correct position as can be judged by the eye. 
Using a plumb-bob, drop a plumb-line from the center of 
the right-hand side of the pulley D, and move the machine 
until the center of the back side of the pulley F touches the 




8G8 



APPLIED MECHANICS. 



plumb-line. In case it should not be convenient to move 
the machine, shift the pulleys instead. If it be desired to 
run the belt in the opposite direction to that indicated by 
the arrow in Fig. 387, shift the machine carrying F to the 
left until the center of ih^ front side of the pulley 7^ touches 
a plumb line dropped from the center of the left-hand side 
of the pulley D\ that is, from the point e. 

1508. There is the objection to a quarter-turn belt 
that, when the angle at which the belt is drawn off the 

pulleys is large, the belt 
is strained, especially at 
the edges, and it does 
not hug the pulleys well. 
Small pulleys placed 
quite a distance apart, 
with narrow belts, give 
the best results, from 
which it follows that 
quarter-turn belts, like 
the foregoing, are not 
well suited to transmit 
much power. Fig. 388 
shows how the arrange- 
ment can be improved 
by placing a guide pul- 
ley against the loose 
side of the belt. The 
^^^- ^s- driver D revolves in a 

left-handed direction, making a b the driving or tight side 
of the belt. To determine the position of the guide pulley, 
select some point in the line a b, as d\ draw lines c d and 
e d\ the middle plane of the guide pulley should then pass 
through the two lines. Looked at from a direction at right 
angles to pulley /% line -^ d coincides with a b\ looking at 
right angles to pulley D, line e d coincides with a b. 

1 509. A third form of belting for connecting two shafts 
at right angles consists of pulleys placed as in Fig. 389. 




APPLIED MECHANICS. 



869 



In general, it is to be preferred to the quarter-turn belts. 
D IS the driver. The belt passes around the loose pulley Z, 
and up again around a loose pulley on the driving shaft back 




a 



Fig. 389. 

of D. It then goes down, around i% which is fast upon the 
shaft a a^ and finally up again and around D. Since the loose 
pulleys revolve in a direction opposite to that of their shafts, 




Fig. 390. 



their hubs should be long, 
must be of the same size. 



The two pulleys on each shaft 
It is evident that either F ov D 



can be the driver, and run in either direction. 



870 



APPLIED MECHANICS. 



It is to be observed that while a quarter-turn belt can be 
used with the shafts at an angle other than a right angle^ 
the last arrangement cannot. 

1510. In Fig. 390 is shown a method of connecting the 
shafts when it is not possible to put the follower F directly 
under D. The guide pulleys G, G' must be so placed that 
the belt will lead correctly from the point a into the middle 
plane of the guide pulley G\ from b into the middle plane 
of D^ and so on around. By twisting the belt at ^, the same 
side will come in contact with all the pulleys; this is a 
desirable arrangement. 

1511. We now come to a case that is different from 
the preceding, in that the shafts A and B, Fig. 391, would 




Fig. 391. 



intersect, if long enough, as, for example, where shafting 
running on two sides of a room is to be connected. Guide 
pulleys, like those in the figure, termed mule pulleys, 
are used. As their planes are horizontal, means must 
be provided to prevent the belt from running off at 
the bottom. Sometimes this is done by simply crowning 
the pulley, and sometimes by putting flanges on the 
lower sides. 



1512. Other Examples of Belt Transmission. — 

Guide pulleys are sometimes used to lengthen the belt 
between two shafts which are too close together to be 



APPLIED MECHANICS. 



871 



connected directly, or it may be that it is not possible to get 
two pulleys in the same plane. Fig. 392 shows an arrange- 
ment of this kind. The diameter of the guide pulleys should 
equal the distance between 
the planes of D and F. 
With the guide pulleys 
arranged as shown, the 
belt will run in both di- 
rections. It is more con- 
venient, however, to place 
them on one shaft. In 
that case their axes would 
be on the line O 0\ G' 
would have to be in the 
line C K^ and G in the line 





Fig. 393. 



A B. Then the belt would be delivered 
from D into the middle plane of G\ 
and from G into the middle plane of 
F. The belt would run in only one 
direction, however. 

1513. A device for connecting two 
horizontal shafts making an angle with 
each other is given in Fig. 393. It can 
be used where a quarter-turn belt would 
not work successfully. The guide pul- 
leys turn in the same direction, Avhich 
is a convenience, because they can then 
be mounted on one shaft, turning in 
bearings at the ends, and the belt will 
run in either direction. 



872 APPLIED MECHANICS. 

WHEELS IN TRAINS. 

1514, The principles relating to the velocity ratio of 
pulleys connected by belting" apply to any series of wheels 
arranged in a train. Where gears are used, however, the 
relative proportions of the wheels may be stated in terms of 
the numbers of teeth, instead of in terms of their diameters, 
if desired. Moreover, in any one train, it is only necessary 
that the proportions of each pair of wheels be stated in the 
same terms; different pairs may be given in different terms. 

For example, take a train of four axes or shafts and six 
wheels, as in Fig. 394. Four of the wheels are gear-wheels. 




Fig. 394. 

represented by their pitch lines, and two are pulleys con 
nected by a belt. Let D^^ D^^ etc., denote the drivers, and 
i^„ F^^ etc., the followers, N the number of revolutions of 
Z^j, and n the number of revolutions of F^. 

Suppose D^ to be 40 inches in diameter and F^ 35 inches ^ 
D^ to have 54 teeth, and F^ 60 teeth; D^ to be 1 foot in 
diameter, and F^ 2 feet. What is the speed of F^ if A^= 100? 

Placing the product of the drivers X the speed of the first 
shaft = the product of the followers X speed of the last shaft, 
100 X 40 X 54 X 1 = ?2 X ^5 X 60 X 2. Hence, 

100 X 40 X 54 X 1 ^, „ . . 

n = ^ — = 514 rev. per mm. Ans. 

35 X 60 X 2 ^ ^ 

It is evident that the arrangement of the drivers and 

followers is indifferent, and that they may be interchanged 

among themselves. It should be noticed, however, that if 

the diameter of one driver be given in inches, the diameter 

of its follower must be given in inches, also, etc. 

1515. Direction of Rotation. — Axes connected by 
gear wheels rotate in opposite directions, as though con- 



APPLIED MECHANICS. 873 

nected by a crossed belt. Hence, in a train consisting solely 
of gear wheels^ if the number of axes be odd^ tJie first and last 
w lie els will revolve in the same direction ; if the nnmber be 
even^ they will revolve in the opposite direction. 

It is evident that a pinion working in an internal gear is 
an exception to this rule. It will turn in the same direction 
as the gear. 

1516. Idle ^Wheels. — In the train, in Fig. 394, shaft 
No. 3 carried two gears, one being driven by gear Z^^ ^^^ 
the other driving F^. Sometimes, however, only one inter- 
mediate gear is used, serving both as a driver and a follower. 
Such a wheel is called an idle wheel, or idler ^ and while it 
affects the relative direction of rotation of the wheels it is 
placed between, it does not affect their velocity ratio. 

The following examples will illustrate this, as well as show 
a few ways in which idle wheels are used. One method of 
arranging the change gears on the 
end of an engine lathe for chang- 
ing the speed of the lead screw, 
which is used to feed the tool in 
screw cutting, is shown in Fig. 




395. Z^j, the driver, receives mo- 
tion from the lathe spindle, and 
i^2, the follower, is on the lead 
screw. The middle wheel, which is 
an idle wheel, acts both as a driver 
and follower, or as D^ and F^. i^^<^- s^^- 

Letting iV^, represent the number of revolutions of D^ and 
n^ of ^2, v/e have, from formula 137, N^X F)^X D^ 
= n^ X F^ X F^. But, as F^ and F^ represent the same wheel, 

N D 
they have the same value, and n^ = — ^r— \ That is, the 

2 

speed of F^ is exactly the same as though no idle wheel was 
used. 

1517. To change the speed of the lead screw, a differ- 
ent size wheel is put on in place of F^, or F^, or both. The 
idle wheel turns on a stud clamped in the slot in the arm A, 



874 



APPLIED MECHANICS. 



This stud can be moved in the slot to accommodate the differ- 
ent sizes of wheels on Z, and to bring the wheel in contact 
with the gear on 6", the arm is swung about L until in the 
right position, when it is clamped to the frame by the bolt B. 
The way in which an idle wheel changes the direction of 
rotation is well shown in Fig. 396, which is a reversing 

mechanism, sometimes 
placed in the head stock 
of a lathe for reversing the 
feed. Here the idler / is 
in contact with D and i% 
making three axes, an odd 
number, so that D and F 
revolve in the same direc- 
tion. Wheels / and /', 
however, are both pinioned 
on the plate P, which can 
swing about the axis of /% and are always in contact. More- 
over, as plate P swings about its center, I must necessarily 
remain in contact with F. If the handle H be moved to 
the right, the plate will be revolved to the right, by means 
of the pin working in the slot in the lever L. The two 
idlers will take the dotted positions shown, and D will drive 
F through both of them. The number of axes will then be 
even, so that D and F will revolve in the opposite direction. 




Fig. 396. 



1518. Fig. 397 shows 
an idle wheel is used, 
known as a knee- 
joint. If shaft A is 
fixed and . shaft C is 
compelled by some ar- 
rangement, not shown, 
to move along a path, 
as in n^ and at the same 
time it is desired to drive 
C from A, the connec- 
tion can be made as 



still another device in which 




Fig. 397. 



APPLIED MECHANICS. 8?5 

shown. Wheels D and F are fast to shafts A and C, and 
the idler / is supported by the links L and K^ the ends of 
the links being loose upon shafts A^ B, and C. Link L 
serves to keep/? and /at the right distance from each other, 
and K serves the same purpose for / and F. The dotted 
lines show another position of the wheels. Among other 
applications of the knee-joint is one used to drive the rollers 
which deliver the ingot in a rolling mill to the rolls. These 
latter are ranged above one another in pairs, and the rollers 
have bearings in an iron frame, called the roll table, that is 
raised or lowered by hydraulic pressure, thus bringing the 
ingot in line with any pair of rollers. A familiar example of 
this train is seen in routing machines, pulleys and belts be- 
ing used, however, in place of gears. The knee-joint is in 
reality an example of an epicyclic train, a description of 
which is to follow. 

ENGINE LATHE TRAIN. 

1519. Some of the best examples of wheels in trains 
are to be found in engine lathes. Fig. 398 represents the 
head-stock of an engine lathe, the spindle vS turning in the 
bearings, as shown, and having a face-plate P, and a "cen- 
ter " on the right end for placing the work. The lead screw 
Z, used in screw-cutting, is connected with the spindle by 
the train of gears on the left, which will be described later. 
The back gears F^ and D^^ on the shaft in n^ have been 
drawn above the spindle for convenience of illustration, 
instead of back of it, where they are really placed. 

1520. Back Gear Train. — It is important to keep 
the cutting speed within limits that the tool will safely 
stand. For turning work of different diameters and ma- 
terial, therefore, the spindle must be driven at different 
speeds. This is accomplished by means of the cone pulley 
marked (T, driven by a similar one on the countershaft by 
means of a belt, and by the back gears. 

The gear F^ is fast to the spindle and always turns with 
it. The cone C^ however, is loose on the spindle, but can 
be made to turn with it by means of a lug, or catch, operated 



876 



APPLIED MECHANICS. 



by a nut under the rim of F^. When the nut is moved out 
from the center, the lug engages with a slot on the large 
end of the cone. The cone will then revolve with the 
spindle, and as many changes of speed may be had as there 




Fig. 398. 

are pulleys on the cone. As ordinarily constructed, how- 
ever, the cone alone does not give a range of speed great 
enough to include all classes of work, nor is the belt power 
sufficient for the larger work and heavier cuts. It makes 
the lathe more compact and satisfactory to construct the 
cone for the higher speeds and lighter work, and to obtain 
the speeds for the heavier work by means of back gears. 
Referring to the figure, it will be seen that the back gears 
are connected by the sleeve .$", and so turn together, F^ 
meshing with D^, and D^^ which is loose upon the spindle 
but fastened to the cone, with F^. To get the slower speeds, 
the nut mentioned before is moved in towards the center of 
F^^ disengaging the gear from the cone, which latter is now 
free to turn on the spindle. Hence, if the back gears are in 
mesh with the gears on the spindle, the belt will drive the 
spindle at a slower speed through the cone and the train 
D F D F, 

The back gears cannot remain in gear when the cone and 
gear F^ are connected ; otherwise, the lathe would not start, 



APPLIED MECHANICS. 877 

or teeth would be broken out of the wheels. To provide for 
throwing them in and out of gear, as required, the rod on 
which the back gears and sleeve s revolve is provided with 
eccentric ends at 7n and 7Z, fitting in bearings in the frame. 
By turning the rod part way around by means of the handle 
H the gears can be thrown either in or out of gear. 

1521. A lathe is spoken of as running back geared 
when the back gears are in, and as being in single gear when 
they are out of gear. This same arrangement, or a modifi- 
cation of it, is used on upright drills, boring mills, milling 
machines, and other machine tools. 

1522. Screw Cutting. — One of the chief uses of 
engine lathes is for screw-cutting. In Fig. 398, the screw- 
cutting mechanism is driven from d^, which is fast to the 
spindle. This connects with the lead screw Z, through 
/*j, d^^ the idler /, andy^- To cut a screw-thread, the work 
is placed between the centers of the lathe and made to turn 
with the face plate by the dog B, clamped to the end of the 
work. The spindle runs towards the operator, or in a right- 
hand direction, as looked at from the outer end of the head- 
stock, and the carriage, tool post, and tool, all of which are 
here represented by the pointer, are moved by the lead 
screw, on shears parallel with the axis of the spindle. 

When cutting a right-hand thread, the tool must move 
from right to left, and from left to right when cutting a left- 
hand thread. Hence, to cut a right-hand screw with the 
gearing as arranged in the figure, a left-hand lead screw 
must be used. To cut a left-hand screw another idle wheel 
should be inserted in the connecting train, arrangements 
for which are usually provided. To cut screws of different 
pitch a set of gears is always furnished, any of which may 
be placed on the stud T, or the end of the lead screw. The 
method of adjustment to accommodate the different size 
gears by. means of the vibrating slotted arm A has already 
been explained. 

1523. Suppose the lead screw to have six threads to 
the inch, or a pitch of ^ inch, and let it be required to 



878 APPLIED MECHANICS. 

cut a screw of six threads per inch. If the gearing is such 
that the lead screw turns once while the spindle makes one 
turn, it is evident that the tool will cut six threads to the 
inch. If it is required to cut more than six threads to 
the inch, however, the lead screw must turn slower than the 
spindle; if less than six threads, it must turn faster than 
the spindle. 

Now, the ratio between the speed of the spindle and the 
speed at which the lead screw must turn for cutting a cer- 
tain number of threads per inch is simply the ratio between 
the number of turns made by the spindle and the number 
of turns made by the lead screw while the tool moves along 
one inch. This is evidently equal to the ratio of the num- 
ber of threads per inch to be cut to the number of threads 
per inch on the lead screw. Thus, if 10 threads per inch 
are to be cut, and the lead screw has six threads per inch, 
the spindle must turn 10 times while the lead screw turns 
six times ; if 4^ threads are to be cut, it must turn 4|- times 
to six turns of the lead screw. 

The problem, then, is to determine what size gears should 
be placed on the stud and lead screw to give the latter the 
right speed for any case. Here, as heretofore, the principle 
must hold that the speed of the first driver, multiplied by 
each driver, equals the speed of the last follower, multiplied 
by each follower. From this and from what has gone 
before we have, therefore, that the number of threads per 
inch of the screw to be czit^ multiplied by the num,ber of teeth 
of each driver^ equals the number of threads per inch of the 
lead screw ^ multiplied by the number of teeth of each follower. 

It is a simple matter, therefore, to find the ratio between 
the gear on the stud and the gear on the lead screw, and 
from that to determine what gears will be suitable. 

1524. The process to be followed will be made clear by 
the following : 

Example. — Let the number of teeth in the different wheels in Fig. 
398 be as follows: ^i, 30; /i, 60; /, idle wheel. Assuming L to have 
six threads per inch, what change gears should be used to cut 4 threads 
per inch ? 



APPLIED MECHANICS. 



879 



Solution. — By formula 137, letting /V and // be the number of 
revolutions of the spindle and lead screw, respectively, we have Ny,di 

Xd^^n X/i X/2, or 4 X 30 X ^2 = 6 X 60 x/^. Hence, d^ = ^^/^^/" 
= 3/2. 

That is, the number of teeth of the gear on the stud (= d^) 
must equal the number of teeth of the gear on the lead 
screw, multiplied by three. In other words, the gear on the 
stud must have three times as many teeth as the gear on 
the lead screw. 

In like manner we have 

12 

for 5 threads, d^ = -z-/,; 



for 6 threads, d^ = 2/,; 
for 7 threads, d^ = -ly-fi* 

etc., etc., etc. 

For four threads we might use wheels of 84 and 28 teeth ; 
for 5 threads 84 and 35, etc. In this way, a table might be 
calculated and arranged in columns, as below : 



Threads per Inch. 


Gear on Stud. 


Gear on Lead Screw. 


4 


84 


28 


5 


84 


35 


6 


84 


42 


7 


84 


49 


etc. 


etc. 


etc. 



Such reference tables are always provided with lathes, and 
the gears are generally so chosen that the smallest number 
possible will have to be furnished with the lathe to cut the 
range of threads desired. Thus, in the above table, the 
gears were chosen so that one with 84 teeth would serve for 
cutting several different threads. 

1525. Many lathes adapted for screw-cutting are not 
provided with the stud T, the gear d^ being keyed directly 
to the end of the spindle 6". 



880 APPLIED MECHANICS. 

In such a case, if it is desired to find whether the lathe 
will cut a certain thread not marked on the plate which is 
usually attached to the head-stock, write the number of 
threads to be cut as the numerator of a fraction and the 
number of threads on the lead screw as the denominator -, 
multiply both numerator and denominator by some number, 
and ascertain whether any of the gears in stock have the 
number of teeth corresponding to the results obtained; if 
not, multiply again by some other number, etc. Thus, sup- 
pose the lead screw has 8 threads per inch and it is desired 

to cut a thread which shall have 7|- threads per inch. 

71 
Forming the fraction gives ~. Multiplying both numera- 

8 

45 
tor and denominator by 6 gives — . Should any of the 

gears have 45 and 48 teeth, the 45-tooth gear should be put 
on the lead ccrew and the 48-tooth gear on the spindle, 

1526. Again, when the lathe has a stud, the gears </^ 
and y*j are never changed, and the required change gears 
may be found by multiplying the fraction formed as directed 
above by a second fraction whose numerator shall be the 
number of teeth in d^ and denominator the number of teeth 
in^j, and then proceeding as before. 

Thus, suppose it is desired to cut a thread having 19 

threads per inch, the lead screw having 8 threads ; the gear 

19 24 
djy 24 teeth, and the stud pinion y^, 48 teeth. Then, "^ X-j^ 

= -— ; a.nd -— X T = Ti- 1 hence, a gear having 76 teeth 
lb lb 4 04 

placed upon the lead screw, and one having 64 teeth placed 

on the stud, will cut a screw having 19 threads per inch. 



REVERSING MECHANISMS. 
IS^T, It is frequently required to design machinery, 
parts of which must have a reciprocating motion alternately 
in opposite directions. The various crank motions are some- 
times employed for this purpose, but in many cases their 



APPLIED MECHANICS. 



881 



use is inadmissible, as they produce a variable motion, and 
for other reasons are inconvenient. It is also often necessary 
that machinery be constructed to run in either direction, at 
will. All such cases require the use of some reversing 
mechanism other than those previously described. A few 
additional examples will now be explained. 

1528. Mangle Gearing.— Figs. 399 and 400 show the 
principle of mangle gearing, so named from its use in 




Fig. 399. 

mangles for pressing clothes. Its principal use, at the present 
time, is for actuating the tables of printing presses. In Fig. 
399 is a rack R, shown in cross-section, and so constructed 
that the gear-wheel W can run in mesh with it, eithei on 
top in position W, or underneath, as at W. The racl^ is 
fastened to a frame i% also in section, attached directly to 
the table of the press. The object is to drive the table back 
and forth in a direction to and from the reader. The method 
will be understood by reference to Fig. 400, which is the 
rack and gear of Fig. 399, looked at from the left. As the 
frame F would obstruct the view of the other parts, it is 
omitted. A longitudinal section of the rack is shown. 

Suppose the gear IV to be on top of the rack, and to 
turn uniformly in the direction indicated. The rack will be 
driven to the right with a velocity equal to that of the pitch 
circle of the gear, until the roll A comes under the gear. 
One tooth of the gear is here omitted, leaving a large space 
which is rounded to fit the roll. From this point the gear 
D. O. ill.—Ui 



882 



APPLIED MECHANICS. 



will gradually drop down in the line in n. At position 2 it 
will have made a quarter-turn^ and the rack will have 
moved a distance r ( = radius of the gear) to the right. At 
position S the rack will be at its former place, and further 
rotation of the gear will drive it to the left at a uniform 
speed. The motion of the rack is harmonic during the 




Fig. 400. 

reversal. To place it under full control of the gear at the 
points of reversal, ** shoes " .S", S are fastened securely to the 
frame F. The inner curves of these are concentric with A 
and D^ respectively, and serve as surfaces for the roll C on 
the gear to roll upon. 

It is evident that the length A /?of the rack must be equal 
to the circumference of the gear, or some multiple of the 
circumference, so that the wide space will come in contact 
with the rolls at each end. If, however, two spaces are 
cut diametrically opposite to each other, length A D may 
be some multiple of the half circumference. The total length 
of the stroke of the table is A D^ plus the diameter of gear. 

1529. Referring again to Fig. 399, the driving 
mechanism consists of tight and loose pulleys T and L on 
the shaft running in bearings B^ B. (9 is a universal joint 
which allows the gear to rise and fall. iT is a square block 
loose on the gear shaft X. It slides in a slotted guide G, 
and constrains the gear to move in a vertical straight line at 
the points of reversal 



APPLIED MECHANICS. 



883 



The advantages of this gearing are that, neglecting the 
irregular motion of the universal joints, a uniform recipro- 
cating motion of any length of stroke may be given the table 
with a gradual reversal. The disadvantage is that, where 
once constrained, the stroke cannot be adjusted. 

1530. Clutch Gearing. — A reversing mechanism 
often used, especially when the reversal must take place 
automatically, is shown in Fig. 401. It consists of three 




Fig. 401. 

bevel gears, of which D is the driver, fast on the driving 
shaft 5. F^ and F^ are continually in contact with D^ and 
are loose upon shaft in n. A sleeve C can move endwise 
upon in ;2, but is compelled to turn with it by a key a pro- 
vided in the shaft. When in the position shown, the notches 
clutch with /^j, and inn turns; when in mid-position, m n will 
not turn, and when thrown to the right, the direction of 
rotation will be reversed. 

1531. If the reversal is to be automatic, some provision 
must be made to insure that after C has been pushed out of 
contact on one side, it will be thrown in contact with the 
other gear. One way of doing this is shown in the figure. 
The lever L is pivoted at (9, and at the other end is forked 
to embrace (7, a roller on each prong running in the groove 
shown. On the rod R^ which is pivoted to L at one end and 
slides in a guide G at the other end, are two collars d and <?, 
held in place by set screws. Helical springs are also placed 
on R against the inside of the collars. Now, suppose the 
tappet 7", which is free to slide on R^ be given a motion to 



884 



APPLIED MECHANICS. 



the right through mechanism connected with F^^ but not 
shown in the cut. When it reaches the spring, it will com- 
press it until the pressure of the spring on e is sufficient to 
overcome the resistance of the clutch. Further movement 
of 7^ will move C to the right until free of F^^ when the 
energy stored in the spring will suddenly throw C in contact 
with F^. The time at which reversal occurs can be adjusted 
by changing the position of d and e. 

1532. Sometim'es it is desirable to have a slow motion 
in one direction with a " quick return." Figs. 402 and 403 
show two methods that may be 
used for this purpose. In the 
first, the driver D is made cup 
shaped so as to allow a smaller 
driver d to be placed inside. For 
the slow motion we have d driv- 
ing i^2> and, for the quick return, 
D driving F^. In Fig. 403, S is 
the driving shaft, giving a slow 
and powerful motion through the fig. 402. 

worm gearing, shown clearly in the end view. The quick 
return motion is through the gears D^, -F,, D^, and F^. 

8 





Fig. 403. 

1 533. It is evident that these mechanisms are not suit- 
able for quick-running or heavy machinery. For such use, 
two belts are generally employed in place of the gears, one 
open and one crossed. Often these belts are made to shift 
alternately from a tight to a loose pulley, while in other 
cases they are arranged to drive two pulleys on the same 



APPLIED MECHANICS. 



885 



shaft In opposite directions, either of the pulleys being 
thrown in or out by means of a friction clutch, as, for 
example, in lathe countershafts. 

1534. Shifting-Belt Mechanism. — Of the former 
class, Fig. 404 affords an illustration, as applied to a planer 
operating on metal, //"is a section of the table, which is driven 




Fig. 4&4. 

forwards and backwards by the rack and gearing shown. 
The work to be planed is clamped to the table, and during 
the forward, or cutting, stroke a stationary tool removes the 
metal. As no cutting takes place on the return stroke, the 
table is made to run back from two to five times as fast as 
forward, the latter speed being about 18 feet per minute for 
work on cast iron. There are three pulleys on shaft 1. T 
is the tight or driving pulley, and L^ and L^ are both loose 
(see top view). There are two belt shifters A and B^ in the 
shape of bell-cranks, pivoted at a and b^ respectively. A 
crossed belt, moving in the direction of the arrow, runs from 
a small pulley on the countershaft, and is guided by the 
shifter A. This belt drives the table during the cutting 



886 APPLIED MECHANICS. 

stroke. The other belt is **open," runs over a large pulley 
on the countershaft, and is guided by the shifter B. 

The short arms of the shifters carry small rollers working 
in a slot in a cam plate C^ which is pivoted at O. Each end 
of this slot is concentric about (9, but one end has a greater 
radius than the other. As shown, the two shifters are in 
mid-position, both belts being on the loose pulleys. Suppose, 
however, the rod E be pulled to the left. Shifter B will not 
move, because its roller will continue to be in the same end 
of the slot, which is concentric about O ; but the roller on A 
will be pulled to the left by passing from the end of the slot, 
which is of larger radius, to the end of the smaller radius. 
The crossed belt will, therefore, be shifted to the tight pul- 
ley, which will cause the table to run forward until the dog 
D strikes the rocker S^ throwing the cam in the other direc- 
tion. The effect of this will be to first shift the crossed 
belt from the tight to the loose pulley, and then to shift the 
open belt to the tight pulley ; one motion follows the other, 
and thus decreases the wear and tear of the belts. The 
length of the stroke of the table can be varied by changing 
the position of the dogs, which are bolted to a T slot on the 
edge of the table. 

The table is driven as follows: The belt pulleys drive gear 
Z^j, which is keyed to the same shaft; D^ drives F^^ which in 
turn drives D^^ keyed to the same shaft. D^ drives gear /, 
which drives the table by means of a rack underneath it. 
The circumferential speeds of / and /^^ are evidently the 
same, and the speed of the table is the same as the circum- 
ferential speed of the gear D^. The velocity ratio of the 
gearing is generally expressed in terms of the number of 
feet traveled by the belt to one foot passed over by the 
table. 

1535. Suppose the belt pulleys to be 24 inches in diam- 
eter, and to make N revolutions while the table travels one 
foot. Let the diameter of D^ be 3 inches; of F^^ 26 inches, 
and of D^^ 4 inches. The circumference of D^ is 12-J- inches, 
nearly, so that for every foot traveled by the table, D^ 



APPLIED MECHANICS. 887 

12 24 

(and F^) will turn -—r^- = -^ times. To find the number of 

turns made by the belt pulleys for each foot passed over by 
the table, we have, therefore, 

24. 
i\A X 3 = — X 26, or 

75 

This multiplied by the circumference of the pulley = 8.32 
X 2 X 3.1416 = 52^ + feet. That is, the planer is geared to 
run 52i to 1. 



EXAMPLES FOR PRACTICE. 

1. In a wheel train, a pulley A drives a pulley^ by open belt ; B is 
on the same shaft with a gear C, which drives a gear D; a gear £ 
is on the same shaft with D and meshes with a gear K Suppose A to 
be 30" in diameter and B 26", and let C have 90 teeth; D, 80 teeth; 
£, 70 teeth, and /\ 60 teeth. How many revolutions will B make in 
five minutes, if A runs at 45 revolutions per minute ? 

Ans. 340.74 revolutions. 

2. An engine lathe belt is on the third speed, the diameters of the 
steps upon which the belt is running being 9" on the countershaft and 
4" on the lathe spindle. If the back gears are "in," how many turns 
will the countershaft make for one of the lathe spindle, supposing 
gears Fi and F^ (Fig. 398) to have 64 teeth each, and Di and Z>2 , 24 
teeth each ? Ans. 3. 16 revolutions. 

3. In Fig. 398, suppose the stud T^to make 3 turns while the lathe 
spindle makes 4. What change gears could be used to cut 10 threads 
per inch, the lead screw having 6 threads per inch ? 

Ans. The gears should be in the ratio of 4 : 5, the latter being the 
lead screw gear. 

4. In Fig. 404, let the diameter of the pulleys be 30" ; of Z>, , 4" ; ol 
/)a » 4", and of Fi , 22". What is the ratio of belt to cutting speed ? 

Ans. 41.4 + : 1. 



DIFFERENTIAL GEARING. 

1536. Heretofore, in the discussion of wheel trains, it 
has been assumed that the bearings of the various wheels 
remained fixed. Occasionally, however, cases are met witn 
in practice where the wheels not only turn about their own 
axes, but in which one or more of them revolves bodily about 




888 APPLIED MECHANICS. 

some other axis, thus having a compound motion of rota- 
tion and translation. Such combinations are known as 
differential gearing or epicyclic trains. Their action 
is somewhat confusing at first, but can be made to appear 
simple by applying the principle of successive movements. 

1537. The Two-^Wheel Train.— I. —In Fig. 405 
are shown two gears D and F united by an arm A. Sup- 
pose the number of teeth in 
D to be represented by m^ 
and the number in F^ by n ; 
then, when the arm A is fixed, 
and D is revolved once about 
its axis right-handed, i^will be 

Fig. 405. revolved left-handed about its 

axis — times. Denoting right-hand rotation by -}- and left- 

hand rotation by — , gear D turns + 1 times and gear F^ 

times. This is the ordinary case of transmission of 

n 

motion by two gears. 

1538. 11. — Suppose, however, that gear D is fixed in 

position so that it cannot turn, and the arm A is given a -f- . 

rotation about the axis of Dy the gear F then partakes of 

lit 
two rotations, + 1 about the axis of D and -I about its 

own axis. In order to see this more clearly, imagine the 

two gears to be replaced by two friction wheels whose diam- 

111 
eters are in ratio — ; then, it is perfectly clear that, in order 

to rotate the arm, the wheel F must roll on D in the same 
direction, and that the number of times it turns on its axis 

1H 

is — . The only difference in the motions, in so far as the 

n ^ ■■ 

rotation of F on its own axis is concerned in these two cases, 
is that it has a negative rotation in the first case and a posi- 
tive one in the second case. If the two gears are equal, one 
revolution of the arm will cause F to turn once on its own 
axis. This may also be considered in the following manner: 



APPLIED MECHANICS. 880 

{a) Let the wheels be locked, or wedged together, so that 
neither can turn relatively to the other, and let the whole 
combination, D, F, and A, be turned as one body about the 
axis of D once R. H. (right-handed). The arm A has now 
been turned around once R. H., just as was desired; but, in 
doing so, D has been turned once R. H., when, according to 
the conditions, it should have remained stationary. 

(I?) Hence, the next step is to unlock the wheels, hold the 
arm stationary and turn D back one turn L. H. (left-handed) 

7/1 

to where it should be. This will cause /^ to turn • — times 

n 

R. H. After that is done, each part of the combination will 

have been through just the same relative motion that it 

would have had if the conditions had been carried out 

directly. 

The results of the two steps upon D, F, and A can be 

tabulated thus,-}- indicating R. H. rotation and — indicating 

L. H. rotation; 

D FA 

{a) Wheels locked +1 +1 -f- 1 

{b) Arm stationary — 1 -\ 



W ■ 1 + ^ +1 

The algebraic sum of these in the third horizontal row give? 
the total motion of each part. It will be seen from this that 
if D and F have the same number of teeth, F will make two 
revolutions to one of the arms, one being about its own axis, 
and the other about the axis of D. 

Example. — Referring to Fig. 405, suppose D has 50 teeth, 7% 20 
teeth, and A to be turned 10 times L. H. How many turns will f 
make, and in what direction ? 

Solution. — , D F A 

Wheels locked -10 -10 -10 

50 
Arm stationary +10 — IOx-kt: 

/iO 



- (10 + 25) - 10 



/^ makes 35 turns L. H. 



890 APPLIED MECHANICS. 

1539. HI. — Suppose that the shaft on which i^ turns 
were the crank-shaft of a steam engine, and that the gear 
D were keyed to the end of the connecting-rod, the arm A 
being loose on both shafts. Then, one stroke of the piston 
would carry Z^ to a position diametrically opposite, and the 
return stroke would bring it back to its first position; in 
other words, D would pass entirely around /% but without 
turning on its own axis, because of being keyed to the con- 
necting-rod. The result will be that F will turn twice for 
one revolution of the arm when the gears D and F are of 
the same size. Although difficult to explain in a simple 
manner, a little thought will convince the student of the 
truth of this statement, and if he can obtain a couple of 
gears and try the experiment the result will amply reward 
him for the trouble. The result may be arrived at very 
simply, however, by means of the method of analysis used 
above. Suppose the gears to be locked and the whole com- 
bination, including the connecting-rod, to be revolved once 
right-handed. Denoting the connecting - rod by (7, and 
remembering that the connecting-rod as a whole does 
not revolve, we must now return it to its original position 
by giving it a left-hand rotation (the arm A being fixed). 

This causes /> to turn once left-handed and F. — times right- 

' n 

handed, assuming that D has in teeth and F has n teeth. 

Tabulating the results, we have 



Wheels locked 
Arm stationary 


C 

+ 1 

-1 


D 

+ 1 

-I 


F 

+ 1 

m 


A 

+ 1 












1 + '^ 

n 


+ 1 



The algebraic sum shows that for one revolution oi A^ F 



in 



turns 1-1 times and that C and D do not turn at all, which 

n 

is perfectly true, since, when D is keyed to C, it is no longer 

a separate link, but is a part of C. Consequently, if D and 



APPLIED MECHANICS. 



891 



F are of the same size, m = n and F turns twice right- 
handed for one right-handed revolution of the arm A. 

This motion was used by Watt to drive his engine, some 
one else having patented the crank motion. It is known as 
the sun and planet motion ; the fixed gear being the sun 
and the revolving gear the planet. 

1540. Higlier. 
T^lieel Trains. — 

In Fig. 406 is repre- 
sented a three-wheel 
train, derived from 
the two-wheel train 
by inserting the idle 
wheel /. Here, as 
in previous cases, 
the number of turns 
made by the wheel 

F is entirely inde- Fig. 406. 

pendent of the size or number of teeth in the idler. 

Example. — Suppose all three wheels to have 24 teeth. If the arm 
makes — 5 turns about the axis of Z>, how many turns will F make, 
and in which direction ? 

Solution. — D I F A 

Wheels locked — 5 — 5 — 5 — 5 

Arm stationary 4-5 — 5 -f 5 








-10 







The method of procedure is exactly the same as with the 
two-wheel train, and it will be noticed that the action of / 
in this train is the same as that of F in the two-wheel train. 
/% in the three-wheel train, however, does not turn at all. 
The straight arrows in the figure are supposed to be fastened 
to the wheels, so that the action of the train can be seen by 
noticing the directions in which the arrows point in the 
dotted positions. 

1541. The statement that /^does not turn at all is not 
literally true, since it turns once on its own axis and once on 
the axis Q>i D, What the statement really implies is this: 



892 APPLIED MECHANICS. 

The gear F cannot impart motion to an annular gear 
(supposing it to mesh with one) keyed to the shaft of D] 
that is, if an annular gear, keyed to the axis of D, meshes 
with gear F, and is caused to revolve owing to the rotation 
of the arm A^ it- will receive no motion from the gear -F 
owing to the rotation of F on its own axis. It will, in fact, 
be deprived of a part of the rotation it should receive from 
the arm of A owing to this positive rotation of F^ the 

amount of which is represented by — , in being the number 

of teeth in each of the three gears and / the number of 
teeth in the annular gear. Suppose, however, that the 
three gears D, /, and F were to be replaced by the two 
gears D and F, of Fig. 405, both gears being free to turn 
on their axes. Then, representing the number of teeth in 
the annular gear by /, one -j- revolution of the arm will 

cause the annular gear (which we will call L) to rotate 

in 

— Xn 

1 -| = 1 -|- — times. For the annular gear and gear 

/ / 

ifi 
F^ both turn right-handed (^D being fixed) and F turns — 

times; the number of teeth in L which come in contact 

1H 

with corresponding teeth in -F is — X ?2 = in\ hence, Z will 

make a part of a revolution represented by — due to the 

turning of F on its axis, and one revolution due to the rota- 
tion of the arm A^ the whole movement being represented 

111 
by 1-1 . Using our method of analysis and applying to 

the first case, we have, assuming all three gears attached to 
A to be of the same size, and remembering that an annular 
gear always turns in the same direction as its pinion, 

D I F L A 

Wheels locked +1 +1 +1 +1 +1 

Arm stationary — 1 -\-\ — 1 ~"I ^ 

m 
+2 1- - +1 



APPLIED MECHANICS. 



893 



For the second case, with two gears, 



Wheels locked 



Arm stationary 



D 

+ 1 

-l 



F 

-1-1 






L 

+ 1 
in 



A 
1 



X n 



P 



in 

+ 1+ — 
n 



in 



+ 1+7 +1 



It should not be imagined that the gear F imparts a back- 
ward motion to L in the first case above, for it does not ; its 
rotation simply prevents the arm from imparting to L its 
entire motion, which it would do if F were keyed to the arm. 

1542. A four-M^tieel train, one of the wheels being 
an annular wheel, is shown 
in Fig. 407. D is fixed, the 
other wheels all turning. A 
revolves about the axis of 
F^. D^ and F are on the 
same spindle, F rolling in- 
side of D. 

In working out a train of 
this kind, first consider all 
the wheels, including D^ to 
be locked together and to 
turn with the arm, as in fig. 4or. 

previous cases. Then, make the arm stationary, and turn 
D back to the position it should occupy. 

Example.— Let D have 100 teeth; F, 20; Dx, 45, and Fu 35. A 
makes + 10 turns. Required the number of turns made by the pinions. 

Solution.— D T^and Dx Fx A 

Wheels locked +10 +10 +10 +10 

45 




Arm stationary — 10 



-^50 + 



(Sxao) 







- 40 + 74f + 10 

1 543. Differential Bevel Train.— A differential 
bevel train is shown in Fig. 408. It consists of three 
miter wheels, two of which, E and C, are on the shaft 5, the 




894 APPLIED MECHANICS. 

third revolving on the arm A, which is also on S. It is 

assumed that none of the wheels 
can slide endways upon their shafts. 
This train has certain peculiar as 
well as very useful properties. 

Suppose A to be held rigidly in 

one position, and £ and C to be 

loose on the shaft. Then, if C be 

Fig. 408. turned once one way, £ will turn 

once in the opposite direction, the three wheels forming a 

simple train of gears, B being an idle wheel. 

Now, suppose C to be held in one position and A to be 
turned once about the shaft S. E will then revolve twice in 
the same direction as the arm. That it will do so can be 
clearly seen by applying the principle of relative motions, 
as follows: 

ACE 
Wheels locked +1 +1 +1 

Arm stationary —1 +1 

+1-0 +2 

From the above it will be seen that if we consider C the 
driver and E as fixed, A will revolve one-half as fast as C. 

If we assume that the arm and all the wheels are free to 
turn, and if C is rotated one way at a uniform speed and E 
the opposite way at the same speed, B will revolve upon A^ 
but A will remain still. If, however, C is rotated faster 
than E, the arm will move in the direction of (7 through one- 
half the angle gained ; if C turns slower than E^ the arm 
will move correspondingly the opposite way. 

It will thus be seen that a bevel train is capable of a 
variety of combinations that can be applied in a useful way. 
A few practical applications of this and the other differen- 
tial trains will now be given. 



EXAMPLES FOR PRACTICE. 

1. In Fig'. 406, let D have 25 teeth ; /, 30 teeth, and F, 50 teeth. If 
D is stationary and the arm makes 5 turns L. H., how many turns will 
/^make, and in what direction ? Ans. 2^ turns L. H, 



APPLIED MECHANICS. 



895 



2. In Fig. 407, suppose D to be 40" in diameter; F, 10" ; D^ 14", and 
Fu 16". \i D remains stationary, how many turns will Fx make to 
each turn of the arm ? Ans. 4^ turns. 

3. In Fig. 408, suppose E to make 1 turn R. H. (looked at from the 
right) and A, two turns L. H. How many turns will C make, and in 
which direction ? Ans. 5 turns L. H. 



1 544. Differential " Back-Gears."— Upright drills 
for metal work are sometimes provided with arrangements 
for increasing the range of the speeds and driving power 
that are different from the back-gears explained under the 
engine-lathe train. Fig. 409 shows one arrangement for this 
purpose. S is the shaft, and C the cone pulley which is loose 
on the shaft. Z? is a casting, also loose on the shaft, having 




Fig. 409. 

teeth on the inside, thus forming an annular gear. A plate 
/*, carrying the small gear or pinion /, is fast to the shaft. 
On the left-hand end of the pulley hub is another gear F, 
which is fast to the hub of the cone. 

The action is as follows: A pin, on which there is a collar 
and nut, is clamped in the slot S l\n D. The pin projects 
through D so that when it is placed in the inner end of the 
slot it will engage with a corresponding slot in the plate P, 
When it is lowered, however, the pin disengages with the 
plate, but the collar can be made to fall in a slot in the arm 
A^ which is a part of the frame. In the former position D^ 
P, and Fare locked together, so that the shaft must turn 
with the cone. In the latter position D is locked with the 



896 



APPLIED MECHANICS. 



frame and cannot turn, while F revolves with the cone. The 
plate /*, therefore, revolves with the shaft at a reduced speed, 
the arrangement being similar to that in Fig. 407. 

1545. Differential Motion. — Perhaps the most 
important application of the differential bevel train is to be 
found in spinning machinery, where it is necessary to wind 
the partly twisted fiber, or roving, upon bobbins. As each 
successive layer is wound on the bobbin, the latter becomes 
correspondingly larger and must revolve at a reduced speed; 
otherwise, the roving, which is delivered at a constant speed, 
would be broken. 

1546. In Fig. 410, B is the bobbin at the top of which 
the flyer/" is suspended. The roving passes through a hole 
c in the center of the upper part of the flyer, then down 




Fig. 410, 



through the hollow arm f of the flyer and through a hole in 
the presser/ to the bobbin. This gives a uniform twist to 
the roving. The flyer moves up and down the length of the 
bobbin winding on the roving in helical layers. 

The machine is driven by a belt through the pulley P and 
the shaft 5 vS. The first gear L on the shaft drives the flyer 
at a constant velocity through Mand N. The next gear that 
is fast to the shaft is the miter wheel D. This forms part of 
a differential bevel train , the other wheel on the shaft being 
F, which is loose. To improve the running qualities, there 
are two intermediate wheels, /^, /,, instead of one, as in Fig. 



APPLIED MECHANICS. 



897 



408. A gear IV, shown in section and loose upon the shaft, 
serves the purpose of the arm used in Fig. 408. Its action is 
exactly the same as the arm, the wheel being used simply as 
a means for causing it to revolve about the shaft. The 
object of the epicyclic train is to drive the bobbin, by means 
of F and V, at a varyiitg velocity suited to the varying 
diameter of the bobbin. 

This is brought about by the cone pulleys C^ and C^. The 
former is driven by the main shaft through the gears iT, 7", 
and R, and the latter drives W through gear U. Now we 
have D rotated uniformly in one direction, which turns F the 
opposite way. H^has a motion opposite to that of D^ the 
number of revolutions that it makes depending upon the 
position of the belt on the cones. As the belt is moved 
automatically to the left, H^ turns slower, i^ turns slower, 
and hence the bobbin turns slower as the roving is wound 
on. The gearing is such that the bobbin revolves in the 
same direction as the flyer and just enough faster to take 
the roving as it is delivered. 

1547. Hartford \V^ater-^Vlieel Governor. — An- 
other application of 
the bevel train is to be 
found in the water- 
wheel governor, in 
Fig. 411. W^ and W^ 
are two wide-faced 
pulleys driven by the 
belt which passes 
over the guide pulleys 
P^ and P^. W^ is a 
frustum of a cone, 
with its diameter at 
the middle, the same 
as that of IV^, which 
is an ordinary pulley. 
These pulleys give 

motion to the double fig. 41i. 

bevel wheels F and C, both of which are loose upon the shaft, 
D. 0. III.— 15 




898 



APPLIED MECHANICS. 



through the bevel gears M and R, respectively. The inner 
parts of E and C gear with the miter gear B on the arm A^ 
the latter being keyed to the shaft. The outer part of C 
drives the governor G through the gear K. The work of the 
governor is to shift the belt on the cone by means of a fork 
(not shown) near pulley P^. 

The action is as follows: When the water-wheel is running 
at the proper speed, the belt is at the middle of the cone and 
E and C turn equally in opposite directions. Hence, A does 
not move. Should the wheel run too fast, however, the 
governor balls would rise, since the speed of W^ rnust always 
be proportional to that of the water-wheel. This at once 
causes the belt to be shifted to the small end of the cone, 
increasing the speed of E. This, in turn, makes ^, and hence 
the shaft, rotate in the direction of E, and an arrangement at 
the left-hand end of the shaft S S (not shown) lowers the 
water-wheel gate. If the wheel should slow down, the reverse 
operation would take place. 



GEAR A?i^HEELS. 
1 548. Let two wheels with parallel axes be held in firm 




Fig. 412. 



rolling contact by pressure upon their axes, as in Fig. 412. 



APPLIED MECHANICS. 



899 



If one be turned in either direction, and there is no slipping, 
the other will rotate in the opposite direction with a circum- 
ferential, or surface, velocity equal to that of the first, as 
though connected by a crossed belt, and the numbers of their 
revolutions will be inversely proportional to their diameters. 
Assuming wheel A to be 10" in diameter and B 20", B would 

make — = — as many revolutions as A. 



1549. Should slipping occur, however, ^ would make 
less than one-half as many revolutions as yi, if ^ were the 
driver. To obviate slipping, suppose that pieces like «, a, 
Fig. 413, are fastened at equal distances on the peripheries of 
A and B^ and that corresponding grooves like b, b are cut. 
Then, the projections, or teeth, on one wheel will run be- 




FlG. 413. 

tween the teeth on the other, and B will necessarily revolve 
one-half as often as ^. It is important to notice, however, 
that although the number of revolutions of the wheels in the 
latter case will be in the ratio of 2 to 1, the speed from tooth 
to tooth might vary somewhat from this ratio, unless the 
teeth were of a shape that would give a constant velocity 
ratio. This would result in an uneven motion that would 
be undesirable, even though the variation was very slight. 



900 



APPLIED MECHANICS. 



1550. The object, then, in designing the teeth of gear- 
wheels should be to so shape them that the motion trans- 
mitted will be exactly the same as with a corresponding 
pair of wheels, or cylinders, without teeth, and running in 
contact without slipping. 




Fig. 414. 



Such cylinders, are called pitcti cylinders, and are al- 
ways represented on the drawing of a gear-wheel by a line 
called the pitch circle. (See Fig. 414.) The pitch circle 
is also called the pitch line. 



1551. The diameter of the pitch circle is called the 
pitch diameter. When the word "diameter" is applied 
to gears, it is always understood to mean the pitch di- 
ameter, unless otherwise specially stated, as " outside 
diameter "or " diameter at the root." 



APPLIED MECHANICS. 



901 





lOPitch. 




12Piich. 




1552. Circular and Diametral Pitch.— The dis- 
tance from a point on one tooth to a corresponding point on 

the next tooth, measured 
along the pitch circle^ is 
the circular pitch. It 
is obtained by dividing 
the circumference (pitch 
circle) by the number of 
teeth, and is used in lay- 
ing out the teeth of 
large gears, and also 
when calculating their 
strength. 

It would be very con- 
venient to have the cir- 
cular pitch expressed in 
manageable numbers like 
1 inch, |- inch, etc. ; but 
as the circumference of 
a gear is 3.1416 times its 
diameter, this requires 
awkward numbers for the 
diameters. Thus, a wheel 
of 40 teeth, 1 inch pitch, 
would have a circumfer- 
ence on pitch circle of 40 
inches and a diameter of 
12. 732 inches. Of the two, 
it is more convenient in 
the great majority of 
cases to have the diame- 
ters expressed in num- 
bers that can be easily 

^^^., , handled. In order, how- 

lojPitch. , , . - 

pjg_ 415 ever, to have the pitch 

in a convenient form also, a new pitch has been devised, 
expressed in terms of the diameter and called the dia- 
metral pitcti. 



902 APPLIED MECHANICS. 

1553.. The diametral pitch is not a measurement like 
the circular pitch, but a ratio. It is the ratio of the number 
of teeth in the gear to the number of inches in the diameter; 
or^ it is the number of teeth on the circumference of the gear 
for 07ie inch diameter of the pitch circle. It is obtained by 
dividing the number of teeth by the diameter. 

A gear, for example, has GO teeth and is 10 inches in 

ro 
diameter. The diametral pitch is the ratio of 60 to 10 = — r 

= 6, and the gear would be called a G-pitch gear. From the 
definition, it follows that teeth of any particular diametral 
pitch are of the same size, and have the same width on the 
pitch line, whatever the diameter of the gear. Thus, if a 
12-inch gear had 48 teeth, it would be 4 pitch. A 24-inch 
gear to have teeth of the same size would have twice 48 or 
96 teeth, and 96 -f- 24 = 4, the same diametral pitch as before. 

Fig. 415 shows the sizes of teeth of various diametral 
pitches. 

Diametral pitch has also been defined as the number of 
teeth in a gear of one inch diameter, which amounts to the 
same as the definitions above. 

Using for illustration a wheel 10 inches in diameter with 

60 teeth, we have 

, ., - circumference 10 X 3.1416 ^^, . , 

circular pitch = -^^ -. -. — = — = .524 inch. 

^ No. of teeth 60 

T^. ^1-^1 No. of teeth 60 ^ 

Diametral pitch = — ^. = — - = 6. 

diameter 10 

1 554. Other Definitions. — The other necessary defi- 
nitions applying to the parts of a gear can be readily under- 
stood from Fig. 414. The thickness of the tooth and width 
of the space are measured on the pitch circle. A tooth is 
composed of two parts, the addendum, or outside of the 
pitch circle, and the root, which is inside. 

A line through the outside end of the addendum is called 
the addendum circle, or addendum line, and one through 
the inside part of the root is called the root circle, or root 
line. The amount by which the width of the space is 



APPLIED MECHANICS. 



903 



greater than the thickness of the tooth is called the back 
lasli, or clearance. 

1555. Proportions for Gear Teetli. — With gears of 
large size, and often with cast gears of all sizes, the circular 
pitch system is used. In these cases, it is usual to have the 
addendum, whole depth, and thickness of the tooth conform 
to arbitrary rules based upon the circular pitch. None of 
these rules can be considered absolute, however. Machine- 
moulded gears require less clearance and back lush than hand- 
moulded, and very large gears should have less, proportion- 
ately, than smaller ones. The following table of proportions 
that have been used successfully will serve as an aid in decid- 
ing upon suitable dimensions. Column 1 is for ordinary cast 
gears, and column 2 is for very large gears having cut teeth. 
C stands for circular pitch. 

TABLE 34. 





1 


2 


Addendum 


o.30(r 

OAOC 
O.IOC 
OA^C 
0.52C 


0.30(7 


Root - 


0.35(7 


Whole Depth 


0.65(7 


Thickness of Tooth 


0.4956' 


Width of Space 


0.505(7 







The gears most often met with are the cut gears of small 
and medium size, like those, for example, on machine tools, 
which are almost invariably diametral pitch gears. The 
teeth are cut from the solid with standard milling cutters, 
proportioned with the diametral pitch as a basis. The sys- 
tem is also coming into very general use for cast gearing. 
In all diametral pitch gears, the addendum is made equal to 
1 divided by the diametral pitch, and the working depth 
twice the addendum. The end clearance is usually taken 
equal to ^ of the addendum for cut gears, though the Brown 
& Sharpe Mfg. Co. use -^q- of the thickness of the tooth on 
the pitch line as the clearance. The side clearance, or back 



904 APPLIED MECHANICS. 

lash, Is made just enough to give a good working fit, and 
seldom exceeds -^^ of the pitch. 

Using the above proportions, a 4-pitch gear would have 
the addendum = 1 -f- 4, or :^ of an Inch ; the working depth 
would be 2 X i = i inch, and the clearance -J X i =^j Inch. 
The whole length of the tooth would be -J- -{- -jV = H Inch. 
The thickness of the tooth would be ^ the circular pitch, 
nearly. In a 10-pitch wheel the addendum would be -^-^ inch 
and the length of the tooth ^-J inch ; In a 2-|- pitch it would 
be 1 -^-2-^ = f inch and the length ^^ Inch. 

1 556. Sizing Gear Blanks. — It is quite as important 
to be able to solve problems involving the diameter, number 
of teeth, and circular and diametral pitch of gear-wheels, as 
to be able to lay out the correct tooth curves. Several rules 
and examples will be given covering cases likely to be met 
with. 

For convenience, these symbols will be used: 
P = diametral pitch ; 
£> = diameter of pitch circle; 
O D =^ outside diameter; 
C = circular pitch ; 
iV= number of teeth; 

A = distance between centers of two wheels; 
V= velocity — I. e., revolutions per minute. 

When two wheels run together, small letters, like the 
above, will be used for the smaller wheel, the large letters 
applying to the larger wheel. 

The product of the circular pitch of a gear and the diam- 
etral pitch is always the constant number 3.1416. Hence ^ 
to change circular to diametral pitch, divide S.IJ^IG by the cir- 
cular pitch; to change diametral to circular pitchy divide S, 1J/.16 
by the diametral pitch. That is, 

P^ ?:iMi, and (149.) 



C 

L4] 
P 



C^^A^, (150.) 



APPLIED MECHANICS. 



005 



Example. — If the circular pitch is 2 inches, the diametral pitch, or 
/*, = -^-^ — • = 1.571 inches, nearly. If the diametral pitch is 4, the 

. , ^ 3.1416 r-n~j . , 

circular pitch, or C, = — j — = .7854 inch. 

1557. Table 35 gives in the first two columns values of 
circular pitch corresponding to common values of diametral 
pitch, and in the last two columns values of diametral pitch 
corresponding to circular pitch values. 

TABLE 35. 



Diametral 


Circular 


Circular 


Diametral 


Pitch. 


Pitch. 


Pitch. 


Pitch. 


2 


1.571 inches. 


2 inches. 


1.571 


2i 


1.39G '' 


1 <i n 


1.676 


H 


1.257 *' 


If '* 


1.795 


H 


1.142 *' 


1| 


1.933 


3 


1.047 *' 


n " 


2.094 


H 


.898 '* 


ItV " 


2.185 


4 


.785 " 


If 


2.285 


5 


.628 " 


lA '' 


2.394 


6 


.524 *' 


H " 


2.513 


7 


.449 ** 


ItV " 


2.646 


8 


.393 '' 


li " 


2.793 


9 


.349 *' 


ItV '' 


2.957 


10 


.314 '' 


1 


3.142 ^ 


11 


.286 " 


H " 


3.351 


12 


.262 *' 


i " 


3.590 


14 


.224 " 


13 << 
1 6 


3.867 


16 


.196 '' 


f " 


4.189 


18 


.175 '' 


ii " 


4.570 


20 


.157 '' 


f " 


5.027 



The product of the diameter and diametral pitch in any 
gear is equal to the number of teeth, or 



N=DR 



(151.) 



906 APPLIED MECHANICS. 

Also, knowing the number of teeth and the diametral 
pitch, the diameter may be found from the same formula. 

Example. — (a) If a wheel is 30 inches in diameter and 3 pitch, how 
many teeth has it ? 

(d) Of what diameter is a 2^ pitch gear having 20 teeth ? 

Solution.— (a) N= DF = dX^0 = 90 teeth. Ans. 

,,, ^ iV^ 20 40 o . , . 

{p) Z? = -^ = rtT-=-v- = o mches. Ans. 

The outside diameter of a gear equals the pitch diameter, 

1 N 

plus twice the addendum, or O D = D-\-2 X -p. But D = -jj. 

Hence, to obtain the outside diameter, knowing' the diam- 
etral pitch and number of teeth, we have 

Oi? = ^+3xi = ^^^. (152.) 

Example. — A wheel is to have 48 teeth, 6 pitch; to what diameter 
must the blank be turned ? 

Solution.— By formula 152, O D r:^E^lJ^J^- 8.333 inches. 
Ans. 

Example. — A gear blank measures 10^ inches in diameter and is to 
be cut 4 pitch. How many teeth should the gear cutter be set to 
space ? 

Solution.— From formula 1 52, (9 Z> = "t or N=:ODxP — 

2 = lOi X 4 — 2 = 42 — 2 = 40 teeth. Ans. 

To find the diameter, the circular pitch and number of 
teeth being given, we have, from the definition of circular 
pitch, 

^=IM- (153.) 

Example. — What is the diameter of a gear-wheel which has 75 teeth 
and whose circular pitch is 1.675 inches ? 

_, 1.675X75 .. . , . 

Solution.— Z> = ^ ^A^n = 40 inches. Ans. 
o.l41o 

Having given the distance between the centers of two 
gears and their velocities, the formulas for their diameters 
may be derived as follows : 



APPLIED MECHANICS. 907 

From formula 136, V D = v d, or D ='^, but A = 
— ^ — , ov D = '^A — d. Equating values of D^ 

whencevd=2AV—dV, 

ord(v+V) = ^AV, 

2 A V 
and d = -j^^- — , ( 1 54.) 

where V= velocity of the large gear and v that of the small 
gear. 

In like manner, 

V -\- V ^ ' 

Example. — Given the distance between centers of two gears =: 5^ 
inches. What must be their diameters so that the ratio of their speeds 
will be as 3 is to 1 ? 

Solution.— By formula 154, d- '^^^^^^ - 2^ inches. 

1 + o 

By formula 1 55, Z> = ?_X50<_3 ^ g^ inches. Ans. 



EXAMPLES FOR PRACTICE. 

1. {a) How many teeth has a 2^-pitch gear, 4 feet in diameter ? (/5) 
What is the circular pitch of this gear ? ^ j,g \ {^) 120 teeth. 

■ I [b) 1.257 inches. 

2. What is the outside diameter of a gear blank from which a wheel 
is to be cut having 50 teeth 4-pitch ? Ans. 13 inches. 

3. The pitch diameter of a gear is 25 inches. What is its outside 
diameter, supposing it to be 6-pitch ? Ans. 25.333 inches. 

4. A gear blank measures 10.2 inches in diameter and is to be cut 
10-pitch. How many teeth should the gear cutter be set to space ? 

Ans. 100 teeth. 

5. Given the distance between the centers of two gears = 20". 
What must be their diameters so that the ratio of their speeds will be 
as 6 : 5 ? Ans. 18.181 inches and 31.818 inches. 



1558. Law of Tootti Contact. — The pitcli point 

of two gears is the point of contact (7, in Fig. 416, of the 
pitch lines. It is the point at which the line of centers O O' 



908 



APPLIED MECHANICS. 




intersects the pitch circles. The point of contact is the 

point where two teeth touch each other. In order that two 

gear-wheels may have the 
same relative velocities 
at every point as their 
corresponding pitch cy. 
linders, the tooth curves 
O'must be of such a shape 
that at the point of con- 
tact they will both be 
at right angles to a line 
Fig. 416. N N^ Fig. 416, passing 

through the pitch point and point of contact. This line 
is called the common normal to the tooth curves. 

1559. The path of contact is the curve described 
by the point of contact during the entire action of a pair of 
teeth. This curve always passes through the pitch point. 

1560. Angle and Arc of Action. — The angle 
through which a wheel turns from the time when one of its 
teeth comes in contact with a tooth of the other wheel 
until the point of contact has reached the line of centers is 
the angle of approach ; the angle through which it turns 
from the instant the point of contact leaves the line of 
centers until the teeth are no longer in contact is the angle 
of recess. The sum of these two angles forms the angle 
of action. The arcs of the pitch circles which measure 
these angles are called the arcs of approach, recess, and 
action, respectively. 

In order that one pair of teeth shall be in contact until 
the next pair begin to act, the arc of action must be at least 
equal to the pitch. 



THE EPICYCLOIDAL SYSTEM. 

1561. The Tooth Outline.— In Fig. 417, let 6^ and 
O' be the centers of two pitch circles, in contact at the pitch 
point C ; and let a smaller circle, whose center is at o, be 
tangent to both circles at C. Suppose the three centers to 




ArPLlED MECHANICS. 909 

be fixed, and the circles to move in rolling contact with each 
ether in the direction of 
the arrows, the circle o 
carrying a marking point 
E. E will then describe a 
curve E d on the plane of 
the circle (9, and a curve 
^ ^ on the plane of the 
circle O' . N N, the com- 
mon normal of these 
curves, will pass through fig. 4ir. 

the pitch point C, so they are suitable for tooth outlines. 

Note. — In mathematics, the words normal and perpendicular have 
the same meaning. 

It will be observed that the relative motions of the circles 
are the same as though the small one rolled on the outside 
of O' and on the inside of (9. E e is, therefore, the epicy- 
cloid of B B\ and would answer for the face of a tooth of 
0\ while E d is the hypocycloid of ^ A\ and would serve 
as a flank for a tooth of O. In like manner, faces for O and 
flanks for O' could be generated by a circle inside of B B' . 

Since the method of rolling up and using these curves was 
fully described in the subject of Mechanical Drawing, more 
will not be said here concerning the process. 

1562. Intercliaiiy:eable ^Vtieels. — It is not neces- 
sary that the two generating circles used should be of the 
same diameter ; provided that the flanks and faces which act 
on each other are generated by the same circle as in the 
previous case. It is customar}^ however, to use the same 
size circle for faces and flanks of both wheels, and where it 
is desired to make a set of gears, any two of which will run 
together, the same size circle must be used for all. 

1563. In Figs. 418 to 420 are shown the effects of dif- 
ferent sizes, describing circles upon the flanks of the teeth. 
In the first, the circle is half the pitch circle, and the flanks 
described are radial. In the second, with a smaller circle, 
the flanks curve away from the radius, giving a strong 
tooth, and in the tliird, with a larger circle, the flanks curve 



910 



APPLIED MECHANICS. 



inwards, giving a weak tooth, and one difficult to cut. It 
would seem, therefore, that a suitable diameter for the 
describing circle would be one-half the pitch diameter of 






Fig. 418. Fig. 419. Fig. 420. 

the smallest wheel of the set, or one-half the diameter of a 
12-tooth pinion, which, by common consent, is taken as the 
smallest wheel of any set. 

It has been found, however, that a circle of five-eighths 
the diameter of the pitch circle will give flanks nearly 
parallel, so that teeth described with this circle can be cut 
with a milling cutter. For this reason, some gear cutters 
are made to cut teeth based upon a describing circle of five- 
eighths the diameter of a 12-tooth pinion, or one-half the 
diameter of a 15-tooth pinion. 

It is more common practice to take the describing circle 

equal to one-half the di- 
ameter of a 12-tooth 
than of a 15-tooth pin- 
ion, and this is the size 
used in this Course. 

1564. Rack and 
Wheels. — A rack may 
be considered as a 
wheel having an infinite 
diameter. The pitch line 
of a rack is, therefore, 
Fig. 421. a straight line, and for 

every revolution of the wheel the rack will travel a distance 




APPLIED MECHANICS. 



911 



equal to the circumference of the wheel. The construction 
of the tooth is shown in Fig. 421, describing circles for the 
rack teeth rolling on the line B B^ and forming cycloids. If 
both circles are of the same diameter, and are the same as 
used for generating the tooth curves for an interchangeable 
series of wheels, the rack will evidently mesh with any of 
the wheels. 

1 565. Annular, or internal, gears are those hav- 
ing teeth cut on the inside of the rim. The width of space 
of an internal gear is the same as the tooth of a spur gear. 
Two describing circles are used as before, and, if they are of 
equal diameter, the gear will interchange with spur wheels 
for which the same describing circle was used. 

In Fig. 422 is represented an internal gear with pitch 
circle A A^ inside of which is the pinion with pitch circle 
B B, The generating circle O^ rolling inside of B B^ will 




Fig. 422. 



describe the flanks of the pinion, and rolling inside oi A A, 
faces for the annular wheel. Similarly, the corresponding 
faces and flanks will be described by o'. The only special 
rule to be observed in regard to epicycloidal internal gears 



012 



APPLIED MECHANICS. 



is that the difference between the diameters of the pitch circles 
must be at least as great as the sum of the diameters of the 
describing circles. 

This is illustrated by Fig. 423. A is the pitch circle of an 
internal gear, and B of the pinion. Then, for correct action, 
the difference (i^—<^) of the diameters must be at least as 
great as ^, the sum of the diameters of the describing circles. 
To take a limiting case, suppose A to have 36 teeth and B 
24 teeth. A wheel with a diameter equal to D—d^ as shown 
dotted at E^ would, therefore, have 36, minus 24, teeth, or 12 

teeth. In the 12-tooth inter- 
changeable system, this latter 
would be the smaller wheel of 
the series, and the describing 
circles would be half its diame- 
ter. From this it follows that, if 
D—dj the diameter of E^ is not 
to be exceeded by the sum c of 
the diameters of the describing 
^ circles, B is the largest wheel 
that can be used with A. Hence, 
Fig. 423. when the interchangeable system^ 

is used, the number of teeth in the two wheels must differ by 
xt least the nu^nber in the smallest wheel of the set. If we 
wished, for example, to have 18 and 24 teeth, and the gears 
were to interchange, describing circles, half the diameter 
of a 6-tooth pinion, would be used, this being taken as the 
smallest wheel. 




THE INVOLUTE SYSTEM. 

1566. Let O and O' be the centers of two cylinders 
that are a short distance apart, and D E o. cord that has 
been wrapped several times around them in opposite direc- 
tions, as shown in Fig. 424. If the circle D D' be turned in 
the direction of the full arrow, the cord D E will cause the 
cylinder E E' to turn also in the opposite direction, as shown 
by the full arrow, and points on the cord D E will describe 



APPLIED MECHANICS. 



913 



portions of the involute curve. In order to better compre- 
hend this, imagine a piece of paper to be attached to the 
bottom of each cylinder, as shown, and that the width of 
each piece is the same as the distances between the cylinders. 
Now, suppose that a pencil be attached to the cord at E^ 
the point of tangency of the line D E with the cylinder 
E E\ in such a manner that it can trace a line on the piece 
of paper attached to the cylinder E E'^ if a proper motion be 
given to the cord D E, 
Turn the cylinder D D' 
in the direction of the ar- 
row, i. e., rotate it left- 
handed. The point E will 
travel towards the cylinder 
D D' in the straight line 
E D, and gradually di- 
verge from the cylinder 
E E' . During this move- 
ment, the pencil attached 
at E will trace the involute 
curve m^ shown dotted on 
the piece of paper. In the 
same manner, if the pencil 
be attached at D^ and the 
cylinder E E' be rotated in 
the direction of the dotted 
arrow, the dotted involute 
m' will be traced on the 
piece of paper attached to 
the cylinder D U . Sup- fig. 424. 

pose that part of the pieces of paper to the right of the curve 
in and to the left of the curve ;;/' be removed and that E 
E' be rotated until the curve 7Jt takes the position e d\ also, 
that D D' be rotated until in^ takes the position g h, the two 
curves being tangent to each other at k on the line D E. The 
cord D E will be at right angles to both curves in and in' at 
the point of contact. The curves will, therefore, be suitable 
tooth curves, according to the law of tooth contact, MED 




D. 0. IIL~16 



914 APPLIED MECHANICS. 

always passes through the pitch point. To make it do this, it 
is simply necessary to connect the two centers by the line O 0\ 
and through its point of intersection C with E D to draw 
the pitch circles A A' and B B' . Two gears, therefore, 
with pitch circles A A' and B B\ and with involute teeth 
formed from circles the size of D D' and E E\ will have the 
same relative velocity as two pitch cylinders with radii O C 
and O' C in rolling contact. Such gears are sometimes 
called single-curve gears, because a single involute curve 
serves for both face and flank. 

If the centers O and O' should now be moved apart so 
that the pitch circles do not touch, the relative velocities of 
D D' and E E' would evidently remain unchanged, since 
they are connected by the cord E D, The curves described 
by point k would also be the same, because they would still 
be involutes of the same circles. From this, it follows that 
the distance between the centers of involute gears may be 
varied without disturbing their relative velocity or the action 
of the teeth — a property peculiar to the involute system. 

1567. Standard Gears.— In Fig. 424, let a line T T 
be drawn at right angles to O 0\ Then, the angle D C T\ 
made hy E D with T 7\ is called the angle of obliquity, 

and the circles D D' and E E\ from which the curves are 
derived, and which are tangent to E D^ are called the base 
circles. In standard interchangeable gears, based upon 
the diametral pitch, the angle of obliquity is taken at 15°, 
which brings the distance between the base circle and the 
pitch circle at about ^^ the pitch diameter. It would be 
well to use these values for all gears. 

1568. Fig. 425 shows two standard gears, A A and 
B B being the pitch circles, and D U and E E' the base 
circles. T T was drawn through C at right angles with 
O O'y andiViVwas drawn through C, making an angle of 
15° with T T. The path of contact is along the straight 
line N TV, and the distance along N N from a point on one 
tooth to a corresponding point on the next is called the 



APPLIED MECHANICS. 



915 



normal pitcli. The parts of the teeth above the base circle 
are involutes, and the flanks below the base circle are radial. 

1 569. Interference takes place when a part of one 
tooth crowds another at some point during the action, so 
that the gears will not run. To determine whether any 
pair of involute gears will work well together, draw lines 
O D and O' E, Fig. 425, perpendicular to the line of action 
N N. So long as the intersections of the addendum circles 
(shown dotted) and the line of action fall between points E 
and D, as at ^, there will be no interference. If they fall 
outside, as at </, both wheels will interfere, while, if the 

O 




O' 

Fig. 425. 
addendum circle of only one gear cuts the line of action out- 
side, the teeth of that gear will interfere. Interference can 
be avoided by slightly rounding the ends of the teeth on the 
larger wheels, the amount to be determined by drawing the 
teeth in different positions. In the interchangeable system, 
all the gears are made to run with the smallest one of the 
set by giving epicyclodial points to the interfering teeth, so 
that they will work smoothly with the radial flanks. 

1570. The smallest ivheel in the interchangeal)le 
series has J^ teeth, the same number as in the epicycloidal 



916 



APPLIED MECHANICS. 



system. The reason for this is that it is the smallest wheel 
having a contact of the parts of the teeth which are true 
involute curves during an arc of action, equal to the cir- 
cular pitch. Wheels of ten teeth will run together, however, 
although the action is not correct. 

1571. The in- 
volute rack (see 
Fig. 42G) has the 
sides of the teeth at 
an angle of 15° with 
the pitch line, or per- 
pendicular to the line 
of action A^iV. The 
ends of the teeth 
should be rounded to 
run with the 12-tooth 
pinion. 

1572. Internal Gears. — The construction is shown 
in Fig. 427. The obliquity (= 15°) is T C N, and the base 




Fig. 426. 




Fig. 427. 

circles E E* and D D' are drawn tangent to the line of 
action A^ N\ about the centers O and 0\ respectively. 
The addendum circle for the internal gear should be drawD 



APPLIED MECHANICS. 917 

through F, the intersection of the path of contact iVA^' with 
the perpendicular 6^ indrawn from the center of the pinion. 
The wheel will then be nearly, or quite, without faces, and 
the teeth of the pinion, to correspond, may be without 
flanks. If the two wheels are nearly of the same size, 
points c and d will interfere, which can be avoided by 
rounding the corners. 

1573. In General. — Formerly, the epicycloidal sys- 
tem was used almost exclusively, but of late the involute is 
rapidly gaining in favor. Its distinctive features are, the 
adjustability of the centers of the wheel and the great 
strength of the tooth. The chief objection that has been 
raised against involute teeth is the obliquity of action, caus- 
ing increased pressure upon the bearings. Where the 
obliquity does not exceed 15^, however, this objection is not 
a serious one. 

BEVEIv GEARS. 

15'74. In drawings of spur gears, the tooth curves in 
the epicycloidal system are obtained by causing the gener- 
ating circles to roll upon the pitch circles. The tooth 
curves, however, represent curved surfaces perpendicular to 
the plane of the paper and the pitch circles, and generating 
circles represent the ends of cylindrical surfaces which are 
in rolling contact. It may be assumed, therefore, that 
tooth surfaces are generated directly by generating cylin- 
ders rolling upon pitch cylinders. In bevel gearing, the 
pitch surfaces are cones, which, when in rolling contact, 
have their apexes at a common point, and it may be assumed 
that the tooth surfaces are generated by generating concs^ 
rolling upon the pitch cones. 

In spur gearing, the teeth of two wheels bear along straight 
lines, which are perpendicular to the plane of the paper. In 
bevel gearing, the teeth are in contact along straight lines, 
but these lines are perpendicular to the surface of a sphere, 
and all of them pass through the center of the sphere, whicli 
is at the point where the apexes of the two pitch cones meei. 
That this is the case will now be explained. 



918 



APPLIED MECHANICS. 




1575. In Fig. 428, let COD represent a pitch cone, 
the part C D E B being the pitch surface of a bevel gear, 
and let yi O Ch^ the generating cone. If we suppose the 
generating cone to de- 
scribe the tooth surface 
in nop by rolling upon 
the pitch cone, the line n 
<?, representing the outer 
edge of the tooth, will lie 
upon the surface of a 
sphere whose radius is O ,^,^,'^ 

n. For the point n, which 0^^- 

describes this line, is 
always at a fixed distance 
from the center O ; hence, 
every point in the line n o 

is equally distant from fig 428. 

(9, and as, in a spherical surface, every point is equally dis- 
tant from a point within called the center, it follows that 
n o must lie upon a spherical surface. 

1576. To be 
theoretically ex- 
act, therefore, 
the tooth curves 
for bevel gears 
should be traced 
upon the surface 
of a sphere, as 
shown in Fig. 
429. This method 
is not a practical 
oAe, however, 
and would have 
no advantage 
over what is 
known as Tred- 
FiG. 429. gold's approx- 

imation, which is much simpler and is universally used. 




APPLIED MECHANICS. 



919 



By this method, the tooth curves are drawn on cones tangent 
to the spheres at the pitch lines of the gears, as shown in 
Fig. 430. The process is simply to develop or unwrap the 
surfaces of the cones, the unwrapped surfaces being repre- 
sented by A B C and C D E in the figure. The length of the 
arc A B C is equal to the length of the pitch circle A' C, and 
the arc C D E is equal to the pitch circle C E' . The gear 
teeth are then drawn upon the unwrapped surfaces, precisely 




Fig. 430. 

as for spur gears of the same pitch and diameter. This pro- 
cess is fully described in the subject of Mechanical Drawing. 
The teeth, as laid out by Tredgold's method, will vary 
somewhat from the shape of the spherical teeth, though 
usually the variation is slight. The actual error of the system, 
however, is less than this difference, for though the tooth 
curves on each gear may not be the same as those on the 
sphere, the amount of their divergence from perfect curves to 
transmit a uniform motion will be of no practical importance. 



420 APPLIED >.IECI1ANICS. 

\^ORM GEARING. 
ISTT. A "worm is a screw made to mesh with a wheel 
called a ^worm-^vheel, the two forming >vorin gearing, 

or scre^w gearing. Worm gearing possesses the following 
characteristics: 

I. — The velocity ratio depends upon the pitch of the 
screw, i. e., the distance the screw advances in one revolu- 
tion, and not upon the diameters of the pitch cylinders. If 
the ivonn is single-threaded, it must make as many turns as 
there are teeth in the wheel for every revolution of the latter; 
if double-threaded, it will make one-half as many turns. 

11. — The direction in which the worm-wheel turns depends 
upon whether the worm has a right-hand or left-hand thread. 

III. — The end thrust of the screw causes the motion of 
the wheel. 

1578. Form of Teeth.. — Fig. 431 illustrates a worm 
and wheel. It will be noticed that in the longitudinal section, 
taken through the worm, the threads appear to be like invo- 
lute rack teeth. The worm is usually made in a screw-cut- 
ting lathe, and as it is easier to turn the threads with straight 
sides, it is better that they should be of the involute form. 
Involute teeth should then be used on the wheel of a pitch to 
correspond with the threads on the worm. 

1579. Pitch. — The circular-pitch system ?s almost 
universally used for worm gearing, because lathes are seldom 
provided with the correct change gears for cutting diametral 
pitches. It is not so inconvenient, however, in the case of 
worm gearing as with the spur gearing. If the diameter of 
the worm-wheel should come in awkward figures, the diam- 
eter of the worm can be made such that the distance between 
centers will be any desired dimension. The circular pitch 
of the gear must equal the pitch of the worm. 

1580. Close-Fitting Worm and Wheel. — To make 
a close-fitting wheel, a worm is made of tool steel and then 
fluted and hardened similar to a tap. It is almost a duplicate 



APPLIED MECHANICS. 



921 



of the worm to be used, being of a slightly larger diam- 
eter to allow for clearance. This cutter, or hob, is placed 
in mesh with the worm-wheel, on the face of which notches 
have been cut deep enough to receive the points of the teeth 




Fig. 431 



of the hob. The hob is then made to drive the wheel, and is 
dropped deeper into it at each revolution of the latter until 
the teeth are finished. 

Fig. 431 represents a close-fitting w^orm and wheel. The 
pitch circles of each are in contact at P. The outside diam- 
eter D of the worm may be made four or five times the cir- 
cular pitch. The arcs C K and E F are drawn about O and 
limit the addendum and root of the wheel teeth, the distance 
between them being the whole depth of the teeth. End 
clearance is allowed the same as for spur gearing. The 



922 



APPLIED MECHANICS. 



angle A is generally taken at either G0° or 90°. The whole 
diameter of the wheel blank can be obtained by measuring 
the drawing. 

The object of bobbing a wheel is to get more bearing sur- 
face of the teeth upon the worm thread, making the outline 
of the teeth something like the thread of a nut. 




1581. Worm-^Vlieels Like Spur Gears. — When 

worm-wheels are not to be bobbed, there is little to be gained 

by making the face of the wheel concave to fit the worm. 

It is better to construct the blanks like a spur-wheel blank. 

The teeth can then be cut in a straight path diagonally 

across the face of the 
blank to fit the angle of 
the worm thread. 

This angle may be ob- 
tained as follows: Fig. 
432 represents a right' 
angled triangle cut out 
^^^•*^2- of paper and wrapped 

around a cylinder. The hypotenuse of the triangle forma 

the thread and, as the base is 

parallel with the axis of the 

cylinder, the angle of the worm 

thread is the angle 6^ between the 

hypotenuse and the base of the 

triangle. Hcncc^ the tangent of 

the angle of the thread ^= the cir- 

ciiinference of the cylinder^ di- 
vided by the pitch. 

In Fig. 343 is shown a graphi- 
cal method. The lines a a^ b b^ 

etc., are the development of the 

screw. Angle C = angle C is 

the proper angle for the teeth of 

the worm-wheel. The distance 

P' = P, parallel to the axis of the 

screw, is the pitcli. 




APPLIED MECHANICS. 923 

1582* Interference. — The same rules apply here that 
were given under the involute system. When the worm is 
to work with a wheel having a smaller number of teeth than 
it was designed for, interference will occur. It can be 
avoided by rounding the tops of the threads of the worm, 
but it is easier to make the wheel blank somewhat larger 
than is called for by the number of teeth. The teeth Avill 
then have very short flanks, the action being almost entirely 
upon the faces, where interference cannot occur. 




RATCHET ^WHEELS. 

1583. Fig. 434: represents a ratchet wh^el A turning 
upon the pin O. 6^ is a vibrating lever 
carrying the paivl, click, or catch 

By which acts upon the teeth of the 
wheel. As the arm moves back, or 
right-handed, the click lifts and slides 
over the points of the teeth; when it 
returns, the click drops against a tooth 
and carries the wheel with it. pj^ 4,^ 

In case it should be desired to prevent the wheel from 
moving backwards when the click is moving backwards, a 
fixed pawl, similar to D^ would be made to bear on the 
wheel and drop behind each tooth as it passed under. Here, 
D would allow a left-hand, but prevent a right-hand 
rotation of A. 

1584. In Fig. 434, in order that the arm may produce 
motion in the wheel, its vibration must be at least 
sufficient to cause the latter to advance one tooth ; but by 
arranging several clicks in the same lever it becomes pos- 
sible to give a motion to the wheel corresponding to less 
than one tooth for each vibration of the arm. Fig. 435 
shows such a construction, B^ B\ and B" being proportioned 
so that they come into action alternately. Thus, when the 
wheel A has moved back an amount corresponding to one- 
third of a tooth, the click B^ will be in contact with tooth 



924 



APPLIED MECHANICS. 



b\ and, if the arm should then move the wheel forwards a 
distance of at least one-third a tooth and then return to its 

former position, click B" would fall 
behind h\ ready to turn the wheel. 
Thus arranged, a slight motion is 
obtained and comparatively large 
teeth may be used. A neater 
construction than the above 
would be to 
on one pin 
which case a 
be 




put all the clicks 
side by side, in 
wide wheel would 



Fig. 435. 



necessary. 



1585. Reversible Clicks. — In feed mechanisms, such 
as are used on shapers and planers operating on metal, and 
which must be driven in either direction, an arrangement 
like that in Fig. 436 is used. 
Wheel A has radial teeth, and 
the click, which is made sym- 
metrical, can occupy either posi- 
tion B or B\ In order that the 
click may be held firmly against 
the ratchet wheel, its axis is pro- 
vided with a small triangular 
piece, shown dotted, against 
which is a flat end-presser, 
always urged upwards by a fig. 43g. 

spring (also shown dotted). Whichever position B may be 
in, it will be held against the wheel. 

1586. Adjusting ttie Motion. — Ratchet wheels are 
largely used on machines requiring a "feed." In all such 
cases they must be so arranged that the feed can be easily 
adjusted. This is often done by changing the swing of the 
lever C^ Fig. 434, which is usually connected by a rod with 
a vibrating lever, having a definite angular movement at 
the proper time for the feed to occur. This lever is generally 
provided with a T slot, in which the pivot for the rod can 
be adjusted by means of a screw and nut. By varying the 




APPLIED MECHANICS. 



925 



distance of the pivot from the center of motion, either one 
way or the other, the swing of the arm C can be regulated 
and the feed made to occur in either direction desired. 

1587. Another method of adjusting the motion is 
shown in Fig. 437. The wheel turns upon a stationary shaft 
or stud (9, and the end of the shaft is turned to a smaller 
diameter than the rest and is threaded, thus forming a 
shoulder against which an adjustable shield vS can be 
clamped by the nut n. Back of the wheel is the arm, also 
loose on the shaft, carrying the click 7?, which latter should 
be of a thickness equal to that of the wheel, plus that of the 
shield. The teeth of the wheel may be made of a shape 
suitable to gear with another wheel to which the feed motion 
will then be imparted, or another wheel back of and attached 
to the one shown could be used for the purpose. 




C-- 



B''::^ 



B 



8 



^€-i 



[O 



Fig. 437. 

The extreme left-hand position of B is shown at B'. Here 
the click rides on the shield and does not come in contact 
with the teeth of the wheel. When the click comes to the 
right-hand edge of the shield, however, it will drop into con- 
tact at b\ and, if B" is the extreme right-hand position, the 
wheel will be turned through a space corresponding to three 
teeth. If the shield should be turned to the right, a smaller 
number of teeth would be moved each time; if it should be 
turned to the left, a feed of four or more teeth, up to the 
full capacity of the stroke, could be obtained ; while, with 



926 APPLIED MECHANICS. 

the shield in its mid-position, it would carry the click during 
the whole swing of the arm and there would be no feed. 

1 588. Ratchet and Scre^w. — Where ratchets are em- 
ployed in the feed motions of machine tools, they are made 
to operate a screw, which in turn drives the *'head" carry- 
ing the tool. 

Example. — A ratchet having 80 teeth is attached to the end of 
a screw having six threads per inch. If the click is set to move the 
ratchet three teeth for every stroke of the arm, how much "feed' 
would the tool have, supposing it to be moved directly by the screw ? 

Solution. — One turn of the screw would move the tool | inch. But 
for each stroke, the ratchet and, hence, the screw, moves /g^ of a turn, 
and the tool would travel ^^^ x i = lir = • 00625 inch- Ans. 



MACHINE DESIGN. 



INTRODUCTORY. 



THE DESIGN OF DETAILS. 

1 901 . Rules and formulas for designing many of the most 
common details of machines are given in the following 
pages. In some cases these rules are based on consider- 
ations of strength, as developed in the subject of Strength 
of Materials; in others, the wear to which the parts are 
to be subjected has been the principal element in determin- 
ing the given proportions. In all cases, however, the 
practice of successful designers has been followed in prefer- 
ence to mere theoretical principles. 

1902. The first work a young designer is called upon 
to do is usually that of making drawings of details of 
machines, the general plans of which have been developed 
by his superiors. He will be given the leading dimensions 
of these details, and will be required to make a drawing from 
which the pattern-makers, blacksmiths, and machinists can 
finish them ready for their places in the completed machine. 

1903. In most shops such parts as bolts, nuts, screws, 
pipe fittings, etc., and often other simple parts of machines, 
are bought from factories, where they are made in large 
quantities by special machinery. If the shop is a large one, 
there may be a separate department where these parts are 
made according to fixed standards. The designer should 
know what the practice of the shop in this regard is, and in 
all cases make his details to conform with these standards. 
He should also know the kind of material available, the 
methods employed by the shop in working this material, and 

For notice of copyright, see page immediately following the title page. 



1216 MACHINE DESIGN. 

the capacity and principal dimensions of the tools and ap- 
pliances for doing the work, in order that the detail as 
designed may be built in the most economical manner. 
When designing a machine part it is well to keep the proc- 
esses that will be used in making it in mind ; this will often 
prevent constructions that would be very difficult and ex- 
pensive if built with the machinery in use in the shop for 
which the design is made. 

1904. In making designs of details it is always advis- 
able to draw them to as large a scale as can be used con- 
veniently. The scales commonly used for small details are 
full size ; and for larger ones 6 inches, 3 inches, or H inches 
= 1 foot may be used. A scale of 4 inches or 2 inches = 1 
foot should never be used if it can possibly be avoided. 

1905. Remember that the object of a detail drawing of 
a machine part is to show the workman in the clearest 
possible manner how the part is to be made and finished, so 
that it will take its proper place in the completed machine 
and do the work for which it is intended. The designer 
must, therefore, be very careful to make the drawing show 
the form and dimensions of each portion as clearly as possi- 
ble; the drawing should also show plainly the kind and 
quality of material to be used, and the finish, if any, to be 
given the different surfaces. Use sections wherever the 
general views do not show the form with perfect clearness. 

It is well for a designer to imagine himself in the position 
of a man in the shop who knows nothing of the machine, 
and study his drawing carefully to see if anything can pos- 
sibly be lacking that will be required to make the ideas he 
wants carried out perfectly clear. 

1 906. Ordinary dimensions are expressed in feet and 
inches, and the fractions -J-, i, -J, yV> ^^^- > ^^ ^^ inch. Never 
use the fractions -J-, -|, or ^ in dimensions, as the scales which 
mechanics use are not divided in these fractions. The most 
common scales in use by mechanics for ordinary work are 
two-foot rules divided into inches, numbering from 1 to 24; 



MACHINE DESIGN. 1217 

for' this reason many draftsmen give all dimensions less than 
two feet in inches and fractions of an inch, and dimensions 
greater than two feet in feet, inches, and fractions of an 
inch. 

Unless great accuracy is required, decimals are never used 
in giving dimensions. If decimal values are obtained from 
the calculations they are expressed in the nearest ^, -^, or ^^ 
of an inch, according to the degree of accuracy required. 
In particular cases, where extreme accuracy is needed, the 
dimensions may be expressed in decimals; as, for example, 
the pitch of gear teeth. 

THE GENERAL DESIGN OF A MACHINE. 

1907. The methods to be employed in the general 
design of a machine will vary so. much with different con- 
ditions that no fixed rules or methods of procedure can be 
given. The first thing necessary is a thorough knowledge 
of the work the machine must do, together with its location 
and surroundings and the conditions under which it must do 
its work. Keeping these in mind, the designer must apply 
his knowledge of the principles of applied mechanics, 
strength of materials, and the design and construction of 
details, in such a way as to accomplish the desired end in the 
simplest and most direct manner consistent with the con- 
ditions imposed. ' 

All machines consist of different combinations of a few 
simple principles ; and, in order to be successful, the designer 
must become thoroughly acquainted with these principles 
and the relation they bear to each other. A study of 
machines that have been built for similar work is of great 
assistance in suggesting ideas for the new machine. 

In many cases it will be necessary to make more or less 
complete drawings of a number of different plans, before a 
satisfactory result will be obtained. A combination that 
appears feasible at first will be found to be impracti- 
cable when drawn out in detail, and the parts proper 
tioned so as to give the necessary strength. In other cases 
the motion of some part may be found to be limited in such 

D. 0. llL—n 



1218 MACHINE DESIGN. 

a way as to interfere with the proper working of the 
machine. The difficulty or expense of manufacture may 
also make some otherwise good design impracticable. 

1908. In all work, whether designing details or more 
complicated combinations, keep all calculations, notes, and 
sketches in such a form that they can be preserved for 
future reference. Date these notes and give them such 
titles as will be required to make their purpose perfectly 
clear. In this way ideas that may be impracticable for the 
particular case for which they were originally developed can 
be kept for a possible future use ; and the results of many 
hours spent in calculation will be preserved so as to make a 
repetition unnecessary. Some engineering establishments 
provide their draftsmen with books made of manila paper 
bound in board covers; and all calculations, notes, and 
sketches are made in these books instead of on loose sheets 
of paper that will soon be lost. 

1 909. The following practical rules are often neglected 
by inexperienced designers : 

Make all parts that are subject to wear or breakage acces- 
sible for the purpose of inspection^ repairs^ or renewal. 

Provide 'jneans for adjusting all parts that are subject to 
wear. 

Make careful provision for lubrication. 

Use links and rotating pieces for guiding motion in 
preference to slides. 

Use cranks^ levers^ belts ^ and gear-wheels for transmitting 
motion in preference to cams^ screws^ or worm-wheels. 

Wherever possible^ m,ake the motion of all parts positive ; 
that iSf avoid the use of weights or springs for producing 
motion. 

Use through bolts or T head bolts instead of tap bolts or 
studs ^ wherever it can be done. 

1910. Designers are often required to furnish an esti- 
mate of the weight and cost of a machine from the draw- 



MACHINE DESIGN. 1219 

ings. This is done in the following manner : The volume 
of the different details is estimated by means of the princi- 
ples of mensuration ; the weight can then be obtained by 
multiplying the volume of each piece by the weight of a 
cubic unit of the material of which it is composed, as given 
in a table of specific gravities. When the weights are 
known, the cost of the material is easily found from the 
known market values. The time that will be required for 
fitting and finishing the different pieces is then estimated 
and charged for according to the rates paid for that work. 
In this way the cost of the machine may be estimated with 
a degree of accuracy that will depend on the knowledge the 
estimator has of the time required to do different kinds of 
work in the shops, and his skill in making approximate 
calculations of the volumes of irregular-shaped bodies. 



MATERIALS USED IIV MACHINE 
CONSTRUCTION. 

1911. Cast Iron. — This metal, which has already 
been briefly referred to in Art. 1332, etc., is used very 
largely in the construction of machine parts, particularly 
those that must be massive ; for example, the frames and 
beds of engines, lathes, planers, etc. It is not well suited 
for parts subjected to shocks or for parts requiring strength 
and elasticity. 

The great advantage of cast iron is the ease with which 
it may be given any desired form. Shapes that could not 
possibly be forged from wrought iron may be cast with 
comparative ease. The operation of casting is as follows: 
A pattern is first made of the exact shape of the required 
part ; this pattern is usually made of pine, though metal is 
sometimes used when the castings are small and a great 
number are to be made. The pattern is placed in a bed of 
sand or loam, in which it leaves, after being removed, an 
im.pression or cavity called the mold. The melted metal is 
poured into the mold, and, after cooling, the casting is with- 



1220 



MACHINE DESIGN. 



drawn and finished to the required dimensions. Cast iron 
contracts in cooling about one-eighth of an inch per foot in 
each direction; on account of this contraction, commonly 
called the slirinkage, the pattern must be made that 
much larger than the required casting. In practice this is 
always done by using a shrink rule in constructing the 
pattern. The shrink rule is about -J" longer per foot than 
the standard rule. 

1912. A serious difficulty experienced in the use of 
cast iron is its liability to be thrown into a state of internal 
stress, on account of inequality of cooling after being 
poured into the mold. It is a matter of experience that the 
amount of contraction depends upon the size and thickness 
of the casting. In general, thick and heavy parts con- 
tract more than thin ones; consequently, a casting com- 
posed of both thick and thin parts will sometimes differ 
from the form desired. Again, one part of a casting may 
cool and solidify while another part is still in a molten con- 
dition. The contraction of 
the latter must, therefore, 
strain the part already solidi- 
fied. The casting is thus 
thrown into a state of internal 
stress which must to some 
degree reduce its effective 
strength. 

Take, for example, the case 
of a pulley. When the rim 
is thin, but rigid, it is liable 
to contract and solidify first, and the subsequent contraction 
of the arm may induce a fracture, as shown at a^ Fig. 592. 
If, however, the arms set first, the subsequent contraction 
of the rim may cause a fracture, as shown at b. 

1913. When pulleys are cast with thin rims which 
are not rigid, the casting often takes the form shown 
in Fig. 593. The rim is drawn in at the points where 
it joins the arms, because the arms solidify and contract 




Fig. 592. 



MACHINE design: 



1221 



after the rim has set, and the latter, not being suffi- 
ciently rigid to withstand 
the pull of the arms, is 
distorted. 

These internal stresses 
make cast iron an unre- 
liable material for the 
construction of parts re- 
quiring strength; and it 
should be the aim of the 
designer to prevent these 
stresses as far as may be 
by making all parts that 
are to be cast as uniform 
in thickness as possible, 
and avoid having a large 
boss, or hub, appear in a comparatively thin part, or hav- 
ing a very thick part meet a very thin one. 

1914. It is found that, in cooling, the iron crystals 
arrange themselves perpendicularly to the surface of the 




Fig. 593. 




Fig. 594. 

casting. For this reason an inside angle (^, Fig. 594) is a 
source of weakness, the casting having a tendency to break 
through the line vi n. Such corners should be rounded, as 
shown at b^ Fig. 594, in which case the crystalline arrange- 
ment renders the casting much stronger. 



1322 MACHINE DESIGN. 

1915. Chilled castings are made by lining the whole 
or a part of the mold with cast iron which is protected by a 
thin coating of loam. The cast-iron lining is a good con- 
ductor of heat, and the molten iron is thus cooled off quickly, 
or chilled. The sudden cooling of the casting prevents the 
carbon from separating from the iron with which it is in 
chemical combination, and as a result the portion of the 
casting which is chilled is of white, hard iron. Usually, the 
chilling extends to the depth of from -J to f inch ; the interior 
of the casting is of soft, gray cast iron, which is best for 
resisting shocks, while the chilled surface is very hard and 
durable. 

For malleable cast iron see Art. 1337. 

1916. TVrougtit Iron and Steel. — The leading prop- 
erties of these metals have already been given. They are 
used for parts of machines requiring strength and elasticity, 
such as shafts, bolts, piston rods, and connecting-rods of 
engines, etc. 

Wrought iron or steel machine parts are forged or rolled 
to approximately the required shape, and then finished upon 
the lathe, planer, or other tool. Parts may also be cast of 
steel the same as cast iron. When so made they are called 
steel castings. 

1917. Copper. — This metal is used principally for 
making tubes, steam pipes, expansion joints, and similar 
details, for condensers, boilers, engines, etc. It can be 
hammered or rolled into sheets or drawn into wire; it may 
be cast or forged, but can not be welded. The tenacity of 
cast copper is about 21,000 lb. per square inch; of forged 
copper, about 30,000 lb. per square inch. The tenacity of 
copper may be increased by hammering, wire-drawing, or 
rolling, but it is at the same time rendered hard and brittle. 
The toughness may be restored by annealing. 

1918. Bronze, or Gun-Metal. — This is an alloy 
composed of copper and tin, in the proportion of 90 parts 
of copper to 10 parts of tin. The metal has a tenacity of 
about 35,000 lb. per square inch. It is largely used for the 



MACHINE DESIGN. 1223 

bearings of rotating machine parts. The bronze being softer 
than the iron wears more rapidly, and thus lengthens the 
life of the rotating part. The hardness of the bronze may- 
be increased by increasing the proportion of the tin; for 
bearings required to sustain a great pressure the bronze may 
be composed of 8G parts of copper to 14 parts of tin. A very 
soft bronze is composed of 92 parts of copper and 8 parts of 
tin. This quality of bronze is used for making gear-wheels 
which are subjected to severe shocks. 

1919. Ptiosplior-Broiize. — This is made by alloying 
ordinary bronze with from 2 to 4 per cent, of phosphorus. 
It is now being largely used instead of ordinary bronze, and 
is also employed in place of iron and steel in the construc- 
tion of propeller blades, pump rods, etc. The softer phos- 
phor-bronze has a tensile strength of about 45,000 lb. per 
square inch ; the hardest varieties may have a tenacity as 
high as 65,000 lb. per square inch, while hard, unannealed 
wire has, in some cases, a tenacity of 140,000 lb. per square 
inch. 

1 920. Manganese-Bronze. — This is also called white 
bronze, and is an alloy of ordinary bronze and ferroman- 
ganese. It is equal in strength and toughness to mild steel, 
and may be forged into nuts, bolts, rods, etc. It resists the 
corroding action of sea water, and is, therefore, much used 
for propellers. Both manganese-bronze and phosphor- 
bronze are largely used in marine work. 

1921. Brass. — Brass is composed of copper and zinc 
in the proportion of two parts of the former to one of the 
latter. Its tenacity is about 25,000 lb. per square inch. 
Brass is used for condenser tubes and for various fittings, 
such as valves, cocks, etc. 

1922. AVood. — This material is used to a limited 
extent in machine construction; for example, oak and 
lignum-vitae are sometimes used for bearings; beech and 
hornbeam for cogs of mortise wheels; pine, cherry, and 
mahogany for patterns. 



nu 



MACHINE DESIGN. 



FASTENINGS. 



SCREWS, BOLTS, AND NUTS. 

1923. Screws are used in machine construction for 
three different purposes: 

1. As a fastening for clamping or joining parts together. 
2. For the transmission of motion. 3. For producing pres- 
sure. Screws used as fastenings are called bolts. 



FORMS OF SCREW THREADS. 

1924. The V Thread. — Screw threads are usually 
triangular or square in section, the triangular form being 
best for bolts, and the square form best 
for screws transmitting motion. The 
Seller's triangular or V thread, 
commonly called the American thread, 
or United States standard, which is 
used in the United States, is shown in 
Fig. 595. Fig. 596 is an enlarged section 
of the thread. 





Pig. 595. 



Fig. 596. 



The an£-le between the sides of the thread is 60°. The 
distance/ from one thread to the next is called the pitch 
of the screw. As shown in the figure, a section of a single 
thread is an equilateral triangle, the altitude of which is /; 
to form the United States standard thread ^ the altitude 



MACHINE DESIGN. 1225 

of the triangle is cut off from the apex, and the angle at the 
root is filled in to a like depth. Hence, the real depth of 
the thread, Z^, is f the altitude of the triangle; that is, 

t, = lt. 

But t=p cos 30° = .866/; 

hence, t^ = lt = ,^^p. (207.) 

1925. V threads are sometimes cut without the flat top 
and bottom, the section being a full equilateral triangle ; in 
this case they are commonly called sharp V threads. 

1926. The pitch of the thread depends upon the diam- 
eter of the bolt; it may be obtained approximately by the 
following formula, in which d represents the diameter of 
the bolt: 

/ = .24:4/^+.625 -.175^ (208.) 

The diameter d^ at the root of the thread may be found 
by the following formula : 

d^^d-% t^=d-\.Zp. (209.) 
The diameter d^ must always be used in calculating the 
strength of a bolt. 

Letting n represent the number of threads per inch in a 
screw, we have 

n = \, (210.) 

1 ^" 

Consequently, d^~ d '- — . (21 1.) 

Example. — The external diameter of a bolt is If inches. Find the 
pitch, the number of threads per inch, the depth of thread, and the 
diameter of bolt at root of thread. 

Solution. — 

/ = .24 V 1.375 + .625 - .175' = .164". 

;/=—•= = 6, nearly. Use 6 threads per inch, then, p = 

P .1d4 

^ = .167". Ans. 

/, = . 65 / = . 65 X . 167 = //, nearly. Ans. 

^1 = ^ - 2 A = If - gV = 1 A". Ans. 

Table 43 gives the number of threads per inch, diameter 

of bolt at root of thread, and effective area of bolt at root of 

thread. United States standard sizes: 



1226 



MACHINE DESIGN, 



TABLE 43. 



Diameter of 

Screw in 

Inches. 


Number of 
Threads 
per Inch. 


Diameter at 

Bottom of 

Threads in 

Inches. 


Area at 

Bottom of 

Threads in 

Square Inches. 


d. 


n. 


c/i. 


a. 


i 


20 


.185 


.0269 


^ 


18 


.240 


.0452 


1 


16 


.294 


.0679 


■h 


14 


.345 


.0935 


i 


13 


.400 


.1257 


A 


12 


.454 


.1619 


f 


11 


.507 


.2019 


f 


10 


.620 


.3019 


1 


9 


.*731 


.4197 


1 


8 


.838 


.5515 


H 


7 


.939 


.6925 


li 


7 


1.064 


.8892 


1* 


6 


1.158 


1.0532 


H 


6 


1.283 


1.2928 


n 


5i 


1.389 


1.5153 


If 


5 


1.490 


1.7437 


11 


5 


1.615 


2.0485 


3 


^ 


1.711 


2.2993 


2i 


^ 


1.961 


3.0203 


2i 


4 


2.175 


3.7154 


21 


4 


2.425 


4.6186 


3 


3i 


2.629 


5.4284 


3i 


3i 


2.879 


6.5099 


3i 


3i 


3.100 


7.5477 


3f 


3 


3.317 


8.6414 


4 


3 


3.567 


9.9930 


4i 


^ 


3.798 


11.3292 


4i 


3f 


4.027 


12.7366 


4i 


n 


4.255 


14.2197 


5 


2i 


4.480 


15.7633 


5i 


2i 


4.730 


17.5717 


5i 


21 


4.953 


19.2676 


5f 


2f 


5.203 


21.2617 


6 

«■ .L r 


n 


5.423 


23.0978 



MACHINE DESIGN. 



1227 



1927. Tlie Square Thread.— In Fig. 597 is shown a 
screw with a square thread, and in Fig. 598 an enlarged 
section of the thread. As the name im- 
plies, the section of the thread is a square, 
each side of which is one-half the pitch. 

The pitch of the square thread is usually 
taken double that of the triangular thread 






Fig. 597. Fig. 598. 

for the same diameter of bolt; for example, the pitch of a 
square thread on a 2-inch bolt or rod is -J- inch. Approxi- 
mately, the pitch is \ the diameter of the screw ; that is, 



Hence, 



<-5^- 



4* 



Also, / 



d. 



(212.) 
(213.) 

(214.) 



__ p _ d 

The edges of the threads should be very slightly rounded 
off so as to prevent them from being accidentally flattened, 
which would cause a nut to bind upon the thread of the 
screw. When this rounding off is carried far enough, the 
thread takes the form shown in Fig. 599. This thread is 
especially adapted to withstand rough usage. 

1928. The modification of the square thread, shown in 
Fig. 600, is frequently used for the lead screws of lathes. 
The section of the thread, instead of being square, tapers 
slightly from root to point. This taper is given not only 
because a thread of this form is much easier to cut than the 



1228 



MACHINE DESIGN. 



square thread, but because it enables the nut, which is made 





Fig. 599. Fig. 600. 

in two parts, to readily engage or disengage with the screw. 

1929. The trapezoidal 
screiv thread is shown in Fig. 
601. One face of the thread is 
perpendicular to the axis of the 
screw, and the other is inclined at an 
angle of 45°. From the construc- 
tion of the figure it is evident that 
/=/, the pitch. To form the actual 
thread an amount equal to ^ / is cut 
off from top and bottom of the tri- 
angle ; hence, the real depth t^ is | /. 

When this thread is used, it is gen- 
erally for communicating motion, 
or where great resistance without 
any bursting tendency is required. fig eoi. 

The usual dimensions are given by the following formulas, 
in which the letters have the same meaning as before : 

, %d d 
^ = i5-=6-- 

8 * 
For example, supposing the diameter of 




' 10 



(215.) 
(216.) 



the screw to 

2<^ 
be \\ inches, the pitch of a trapezoidal thread is / = -j-r = 

the number of threads per inch is — = — = 

P \ 
d V- 
5, and the depth of the thread is t^ =-Ta ~ Tn ~ '^^" ' 

1930. The relative advantages of the various forms of 
screw threads may be shown by a consideration of the 



?_Xii-i". 
15 ""^ ' 



MACHINE DESIGN. 



1229 




Fig. 602. 



forces acting on the thread. Usually the load on a bolt or 
screw acts in the direction of its axis ; that is, a bolt used 
as a fastening is in tension, while a screw used to produce 
pressure is in compression. In either case the load is carried 
by the reaction between the surface of the thread of the 
screw and the surface of the thread in the nut. Suppose, in 
Fig. 602, the load to be upon the side in I of the thread, and 
let the reaction of the thread at the 
point A be represented by R^ which 
must, of course, be perpendicular to 
m I. This reaction R may be resolved 
into two forces, one, P, parallel to the 
axis of the bolt, and the other, Q^ per- 
pendicular to the axis. Then, P repre- 
sents the portion of the load carried by 
the surface A of the screw, while the 
force Q tends to burst the nut. Now, 
for a given load the force P will remain 
the same whatever the angle of the thread may be. On 
the other hand, the forces Q and R will increase as the angle 
mill decreases. The friction between two surfaces is pro- 
portional to the perpendicular pressure between them. 
Consequently, the greater the angle of a screw thready the 
greater is the friction between the bolt and nut, and also the 
greater is the force tending to burst the nut. 

In the case of the square thread, the angle between the 
sides is zero, and hence there is no force tending to burst 
the nut. The reaction R becomes equal to the load P\ 
therefore, the friction of a square thread is less than that of 
a triangular thread. On the other hand, the triangular 
thread is nearly twice as strong as a square thread. Thus, 
in Fig. 602, the shearing surface of a single triangular 
thread is 7r</^ x distance nl, or T^d^p, nearly, while it will be 
seen by referring to Fig. 598 that the shearing surface of a 

single square thread is tt ^^ x i/ = — -^. 

It follows, therefore, that the triangular thread is better for 
fastenings, and the square thread for transmitting motion. 



1230 



MACHINE DESIGN. 



The trapezoidal thread combines the good features of 
both the triangular and the square threads. It has the 
same shearing section as the former, and the same friction 
as the latter. In this care must be taken, however, to use 
the screw so that the pressure comes on the flat side of the 
thread, for if it is put upon the inclined side the friction 
and bursting force on the nut are both greater than for a 
60° triangular thread. The trapezoidal thread is used in 
the breech mechanism of large guns. 

In England the 'Whit^vortli system of triangular 
threads is in use. The angle of the Whitworth thread is 
55° and the point is rounded instead of being cut flat. 

1931. Multiple - Threaded Screws. — It is plain 
that a nut will advance a distance equal to the pitch of the 
thread for each revolution of the screw. When a screw is 
used to transmit motion, it is often desirable to have the 
nut advance a considerable distance for one revolution, and 
this may necessitate a pitch altogether too large for the 
diameter of the screw. This difficulty is obviated by cutting 
two or more parallel threads, each having the same pitch. 

These screws are termed multiple-tlireaded screws; 
when the screw has two threads, it is called a double- 
tlireaded screw; when it has three threads, a triple- 
threaded screw, and when it has four threads, a 
quadruple-threaded screw. 

In Fig. 603 is shown a single square-threaded screw, and 
in Fig. 604 a double square-threaded screw, both screws 

having the same 
diameter and pitch. 
It is apparent that 
the root diameter of 
the double-threaded 
screw is much larger 
1 than that of the 
single-threaded one; 
it is, consequently, 
stronger, and is, 
therefore, to be pre- 





r 



Fig. 603. 



Fig. 604. 



MACHINE DESIGN. 1231 

ferred. The distance (/j, Fig. 604) between two consecutive 
threads of a multiple-threaded screw is equal to the pitch /, 
divided by the number of threads (2, 3, or 4, according to 
whether the screw is double, triple, or quadruple-threaded), 
and is called the divided pitch of the thread. 

The dimensions of the thread are based upon this divided 

P 
pitch ; that is, t ^=~. Besides being stronger for the same 

pitch than the single thread, the multiple thread has the 
advantage of having a greater wearing surface than the 
single thread. 

1 932. Gas-Pipe Threads. — The rules for the pitches 
and depth of screw threads do not apply to gas-pipe threads, 
since the calculated depth of the thread would in that case 
be greater than the thickness of the pipe. 



1232 



MACHINE DESIGN. 



The following table gives the standard dimensions of 
steam, gas, and water pipes: 

TABLE 44. 



Nominal 


Thickness 


Actual 
Internal 


Actual 
External 


Threads 

T 1 


Pitch of 


Diameter 

• T 1 


in Inches. 


Diameter 


Diameter 


per Inch. 


Threads. 


m Inches. 




in Inches. 


in Inches. 


n. 




i 


.068 


.270 


.405 


27 


.037 


i 


.088 


.364 


.540 


18 


.056 


f 


.091 


.494 


.675 


18 


.056 


i 


.109 


.623 


.840 


14 


.071 


f 


.113 


.824 


1.050 


14 


.071 


1 


.134 


1.048 


1.315 


Hi 


.087 


li 


.140 


1.380 


1.660 


111 


.087 


H 


.145 


1.611 


1.900 


Hi 


.087 


2 


.154 


2.067 


2.375 


Hi 


.087 


H 


.204 


2.468 


2.875 


8 


.125 


3 


.217 


3.061 


3.500 


8 


.125 


H 


.226 


3.548 


4.000 


8 


.125 


4 


.237 


4.026 


4.500 


8 


.125 


H 


.247 


4.508 


5.000 


8 


.125 


5 


.259 


5.045 


5.563 


8 


.125 


6 


.280 


6.065 


6.625 


8 


.125 


7 


.301 


7.023 


7.625 


8 


.125 


8 


.322 


7.982 


8.625 


8 


.125 


9 


.344 


9.001 


9.688 


8 


.125 


10 


.366 


10.019 


10.750 


8 


.125 



1933. Threads may be right-handed or left-handed. 
To determine whether a screw is right or left-handed, hold 
it so that its axis will be horizontal; if the slope of the 
thread (from top to bottom) is from left to right, it is right- 
handed ; otherwise the thread is left-handed. For nuts the 
above rule should be reversed. The threads of screws for 
general use are right-handed, and are so shown in the pre- 
vious figures. Screws having left-handed threads are made 
only for special purposes. 



MACHINE DESIGN. 1233 

STRENGTH OF SCREW BOLTS. 

1934. Usually the stress on a bolt acts in the direction 
of its axis ; that is, the bolt is in tension. 

Let W= load on bolt in pounds; 

St= safe working stress in pounds per square inch; 

a = area of cross-section of bolt at root of thread; 

^= nominal (outside) diameter of bolt in inches; 
d^ = diameter at root of thread in inches. 
Then, if the bolt is in tension, 

W 
W= a St; or^ a ^z-^. (217.) 

The value of the nominal diameter d (corresponding to 
the value of a) obtained from formula 211 may be found 
from Table 43. 

1935. For bolts subjected to a constant tension, Sf may 
be 8,000 lb. per sq. in. More often the tension varies be- 
tween zero and its maximum value ; in this case Sf may be 
taken as 6,000 lb. per sq. in. For cylinder-head bolts, and, 
in general, for bolts used to make a steam-tight joint, St 
may vary from 3,000 lb. for small cylinders to 6,000 for very 
large ones. Ordinarily S^ may be taken as 4,000 or 4,500 lb. 
per sq. in. All the above values are for wrought-iron bolts. 

Example. — Find the diameter of a wrought-iron bolt which is to 
sustain a steady load of 4^ tons. 

Solution. — From formula 217, 

IV 4^X2,000 
" = :S7- 8,000 = 1-1^5 sq.m. 

From Table 43, the value of d lies between If ^ and 1^^ 
The latter value should be taken. Ans. 

1936. For screws transmitting motion, the following 
formula may be used: 

W= 3,000^," = 1,920^' (since < = ^d), 

or^=.0183/F) 

d=.022Si/W\ ' 

D. 0. III.— IS 



1234 



MACHINE DESIGN. 



For screws of this character, the least number of threads 
in the nut that are necessary to prevent excessive wear is 
given by the following formula, in which n^ = the number 
of threads in the nut : 



n, = 



W 



300^: 



w 

= .0052-7-,. 



(219.) 



The above formula applies to square and trapezoidal 
threads, and is based upon the assumption 
that the pressure on the thread per square 
inch of projected area should not be greater 
than 700 lb. per sq. in. 

Example. — A square-threaded screw 1-| inches 
in diameter transmits motion to a load of 4,000 
pounds. What is the least allowable number of 
threads in the nut ? 

Solution. — 




ni = .0052 -^ = 



W .0052 X 4,000 



(li)^ 



= 9i. Ans. 



PROPORTIONS OF BOLTS AND NUTS. 

1937. The dimensions of the nut and 

bolt head are made to depend upon the 
diameter of the bolt. The standard form 
of bolt and nut is shown in Fig. 605. The bolt has a square 
head and hexagonal nut with washer. The washer is used 
to give a smooth seat for the nut to be screwed up against. 
The following proportions are usually adopted; 
Diameter of nut or head across flats, 



D —l^d-\-y for rough work. 
D = l^d-\- -^" for finished work. 

Height of nut, 

h=^ d for rough work. j 

h=^ d — ^" for finished work. J 

Thickness of washer, t = ,15 d. 

Diameter of washer, I^^ = ^i-Dy 



(220.) 



(221.) 

(222.) 

(223.) 



MACHINE DESIGN. 



1235 



The above proportions for diameters D hold for both 
hexagonal and square nuts. The diameter across corners 
D^ may be found from the geometry of the figure. Thus, 
for hexagonal nuts, 



* COS 30' 



D 



86G 



1.73^+. 14'' for rough nuts. ) i^^^ 
1.73^+.07''for finished nuts. ) ^^^^'^ 



For square nuts, 

^ cos 45 
Height of head. 



2. 12^4-. 18" for rough nuts. 
2. 12^+. 09" for finished nuts 



[ (225.) 



Ji' — ^d-\- J-g-" for rough bolts. ) 



(226.) 



Ji' z=z d — Jg-" for finished bolts. 

Example. — Required, the various dimensions of a finished bolt and 
hexagonal nut, the bolt being 1^ inches in diameter. 

Solution. — Diameter of nut across flats = Z> = 1-^ </ + -jf^- = 1^ X li 
+ tV = 2^ in. 



Diameter of nut across corners = D\ = 



D 2.3135 



= 2H in., 



.866 ,866 
nearly. Ans. 

Side of square bolt head = D = 2^^ in. Ans. 
Height of nut =1 h = d — ^^ = 1^ — jV = Itf in. Ans. 
Height of bolt head = k' = d — ^jr = 1^ — -^^ = 1^ in. Ans. 
Diameter of washer = Z>2 = li -C>i = li X 2||- = 3 in. , nearly. Ans. 
Thickness of washer = / = .15 rtr= .15 X li = i in., nearly. Ans. 



1938. 



TVRENCHES. 

The usual forms of solid wrenches are shown in 




-15d tolSd 

Fig. 606. 



Fig. 606, in which that shown at A is used for hexagonal 
nuts, and that at B for sqjiare nuts. The length may be 



1236 



MACHINE DESIGN. 



from 15 to 18 times the diameter of the bolt for which it is 
to be used. In the figures d represents the diameter of the 
bolt. The other proportions are given in terms of the 
diameter across the flats of the nut as shown in the figure. 



FORMS OF BOLT HEADS. 

1939. The ordinary square bolt head has been shown 
in Fig. 605. Other forms are shown in Figs. 607 to 615. 
In Fig. 607, the hexagonal bolt head is similar to a hex- 
agonal nut, and has the same dimensions except that the 
height h' may be less. Usually Ji! ^=z^ d to d. 

Fig. 608 shows a hexagonal head with a collar or flange, 
which is added to give an increased bearing surface. A 




Fig. 607. 



-l^d- 

FlG. 609. 



ISd^ 

Fig. 610. 



Fig. en. 



cylindrical head is shown by Fig. 609, and a hemispherical 
head by Fig. 610. The height h' of the former may be 
from .5d to . Sd; that of the latter is f </. The diameter of 
these heads is as , shown by the figures. Fig. 611 shows a 
bolt head with a hemispherical bearing surface resting on a 
seat of the same shape. This bolt may lean to one side 
while the head will still remain in contact with its seat all 
the way round. 

An eye-bolt is shown in Fig. 612. The cross-section of 
the eye through the hole should equal or exceed the area of 
the bolt. 

That is, referring to this figure, 2 a ^ = ^ t: d^ or 

ad=.3dd\ 

In good practice ^ ^ is at least equal to ^d^. To calculate 
the diameter d^ of the pin passing through the eye, we ob- 
serve that the pin is in double shear, the shearing surface 

being twice the area of it ; or 2 ( —r^ ) = i^ ^x* 



MACHINE DESIGN. 



1237 



The strength of this pin in shear should equal the strength 
of the eye-bolt in tension; therefore, letting 5^ represent 




Fig. 612. 



Fig. 613. 



Fig. 614. 



Fig. 615. 



the safe shearing stress per square inch, and S^ the safe ten- 
sile strength, we have 



or d. 



\ = d\/. 



S, 



25, 



5 
But the ratio -^J■ is usually about 1.25; 



hence. 



d^ = d\/.(}25 = .Sdj nearly. 



(227.) 



If, however, the pin is overhung; that is, if there is but 
one of the nibs vS", 5, instead of two, as shown in Fig. 612, it 
will be in a single shear, and 

,/5" 
d' = d\^ ^=l.ld, nearly. (228.) 

Fig. 613 shows the head of a liook bolt. This form of 
bolt is used when it is undesirable to weaken one of the con- 
nected pieces by a bolt hole. The proportions are shown in 
the figure. The countersunk liead is shown by Fig. 614, 
and the T liead by Fig. 615. 



1238 



MACHINE DESIGN. 



1 940. The ordinary method of attaching a bolt to stone- 
work is shown in Fig. 616. The head is long and rectangu- 
lar, and is made jagged with a cold chisel ; the hole is made 




Fig. 616. Fig. 617. 

larger at the bottom than at the mouth. The bolt head is 
placed in the hole and the remaining space is then filled with 
melted lead or sulphur. 

1941. The ordinary method of fixing the foundation 

bolts which fasten an engine bed to its foundation is shown 
in Fig. 617. These foundation bolts have no solid heads, but 
are long rods threaded on one end for a nut, and have a slot 
in the other end through which passes a cotter (7, which 
rests against a cast-iron or wrought-iron washer in. This 
washer and cotter form the head of the bolt. 

The bolt head is within a recess formed in the foundation, 
so arranged as to be accessible. The area of the washer 
bearing against the foundation, multiplied by the safe com- 
pressive strength of the material of the foundation, should 
be equal to the tension carried by the bolt. For example, 
the tensile strength of wrought iron is about 20 times the 
compressive strength of brick. Hence, the bearing area of 
a washer resting against a brick foundation should be 20 
times the cross-section of the bolt. 

1942. Various devices are used to prevent a bolt from 
turning while the nut is being screwed up. A common 



MACHINE DESIGN. 



1^30 



method is to make the neck of the bolt next to the head 
square, as shown in Figs. 610, 611, and 613. The holt hole 
is also made square. Another way is to insert a pin a into 
the neck, close to the head, as shown in Figs. 609 and 614. 
The projecting part of the pin fits into a recess cut out to 
receive it. 

FORMS OF NUTS. 

1 943. The common hexagonal nut has been shown in 
Fig. 605. Ordinarily, both hexagonal and square bolt heads 
and nuts are chamfered off 2it an angle of 30° or 40°. Other 
forms of nuts are shown in Figs. 618 to 622. The flanged 




Fig. 618. 



Fig. 619. 



Fig. 620. 



Fig. 621. 



Fig. 622. 



nut. Fig. 618, is useful when the bolt hole is larger than the 
bolt, as it covers the hole and gives a greater bearing sur- 
face. Fig. 619 shows a nut with a splierical bearing 
surface and the seat shaped to correspond. This nut will 
bear upon the seat all around, whether the bolt be perpen- 
dicular or inclined to the seat. A cap nut is shown in Fig. 
620. This form of nut is used to prevent the leakage of a 
fluid past the screw threads. To prevent leakage past the 
seat, the nut is screwed down on a soft, thin copper washer a. 
Fig. 621 shows a round nut, and Fig. 622 an ordinary 
square nut. The round nut is provided with holes in its 
circumference, as shown, and is screwed up by inserting a 
bar b in one of the holes. 



LOCKING NUTS. 

1944. All nuts are slightly loose on their bolts, a small 
clearance being necessary to permit them to turn freely. 
When a nut is subject to vibration it is liable to slack back and 



1240 



Machine design. 



allow the bolt to become loose. To prevent this slacking 
back, various locking arrangements have been devised. A 
common device is the locknut, or jam nut, shown in Fig. 
G23. Two nuts are used, one of which is about half as thicTk 
as the ordinary nut. The load is thrown on the outer nut, 
which should, therefore, be the thicker one. In practice it 
is common to place the thin one on the outside, because the. 
wrench is generally too thick to act on it when placed below 
the other. The jam nut is not always satisfactory as a 
method of locking. 

1945. The nut may be effectively locked to the bolt by 
the use of a set-screw, as shown in Fig. 624. To prevent 
the point of the set-screw from injuring the thread, a piece 
of iron or steel in may be let into the nut. The piece is 
screwed along with the nut, and acts as a shield interposed 
between the set-screw and thread. 

1946. A good method of locking a nut is shown in 
Fig. 625. The lower portion of the nut is turned down, and 




Fig. 6?3. 



Fig. 634, 



Fig. 625. 



a groove is cut in the center of the circular portion. A collar 
is fastened by means of a pin to one of the pieces to be con- 
nected, and the circular part of the nut is fitted into this 
collar. The nut is then bound to the collar by a set-screw 
passing through the latter, the point of the set-screw engaging 
into the groove turned in the nut. The following propor- 
tions have proved very satisfactory, in which d, the diameter 
of the bolt, is taken as a unit. All dimensions are in inches: 



MACHINE DESIGN. 



1241 









(229.) 






1947. Fig. 626 shows a device for locking a nut by 
means of a stop plate. The plate is fastened to one of the 
pieces through which the bolt passes. 

It is so shaped that the bolt may be locked at intervals 
of ^ of a revolution. Suitable proportions for this stop 
plate are shown in the figure, in 
which </, the diameter of the bolt, is 
taken for the unit, ex- 
cept the distance be- 
tween the center of 
the bolt and screw for 
which /?, the diameter 
between the parallel 
sides of the nut, is taken 
for the unit. All di- 
mensions are in inches. 

1948. In Fig. 627 
is shown another form 

of stop plate^ which may be conveniently used when the 
bolts are set in a circle, as, for example, on engine cylinder 

heads. 



1949. In Fig. 
628 is shown a differ- 
ent manner of lock- 
ing the nut. In this 
the nut is sawed half 
way through, and the 
parts connected by a 
small screw. When the 
nut is screwed home 
the small screw is 
tightened, thereby 




Fig. 627. 



Fig. G26. 




Fig. 628. 



greatly increasing the friction between the bolt and the nut. 



1?42 



MACHINE DESIGN. 



1 950. A convenient locking device is Grover's spring 

-washer, shown in Fig. G29. The washer when not held 




Fig. 629. 
down by the nut has the form shown in the lower part of 
the figure ; when the nut is screwed down tightly, the washer 
is flattened out and its elasticity keeps the nut tight on the bolt. 



1 

Fig 

it. 

the 



FORMS OF BOLTS AND SCREIVS. 

951. Bolts. — A stud bolt, or stud, is shown in 

. 630. Each end of the stud has a screw thread cut on 

One end screws into one of the pieces to be connected, 

other carries a nut. a i i • 

A stud having a collar is 

shown in Fig. 631. The col- 
lar may be square or round ; 
it serves as a shoulder against 
which to screw up the stud, 
and, when square, is a con- 
venient place to apply a 
wrench. 

A tap bolt, shown in Fig. 
632, is a bolt not requiring a 
nut. It is screwed directly 




Fig. 630. 



Fig. 631. 



MACHINE DESIGN. 



12 i3 



into one of the pieces to be connected, the head pressing 
upon the other piece. 

Fig. 633 shows a tap bolt having a countersunk head. 
This style of bolt is called a patch bolt, and, as its name 
implies, it is used in making patches in boilers, etc. The 
diameter of the neck of the projection to which the wrench 




Fig. 632. 



Fig. 633. 



Fig. 634. 



is applied is smaller than the root diameter of the bolt, so 
that the projection will break off instead of breaking the 
bolt when too much force is applied to it. 

Fig. 634 shows a bolt having a nut at each end instead of 
a head and nut. 

1952. Screws.— In Figs. 635, 636, and 637 are shown 
different forms of machine screws slotted for a screwdriver. 
Fig. 635 is called a countersink liead screw ; Fig. 636, 




Fig. 635. 



Fig. 636. 



Fig. 637. 



a button bead screiv, and Fig. 637, a fillister bead 

scre^w. When the countersink head screw is used, the hole 
in the piece which is to be held tight is countersunk so that 
the head of the screw is flush as shown. 



1244 



MACHINE DESIGN. 



1953. Set-scrcTvs are screws or bolts which are used 
to press against a piece, and by friction to prevent it from 




Fig. 638. 



Fig. 639. 



Fig. 640. 



Fig. 641. 



moving or rotating relatively to another piece. For ex- 
ample, a set-screw may be screwed through the hub of a 
pulley, and by pressing against the shaft will prevent the 
pulley from turning on the shaft. Various forms of set- 
screws are shown in Figs. 638, 639, 640, and 641. 

Fig. 638 is called a cone-point set-screw ; Fig. 639, a 
cupped set-screw ; Fig. 640, a round pivot-point set- 
screiiV, and Fig. 641, a lieadless cone-point set-screvi^. 

1954. Bolts in Shear. — Usually bolts are in direct 
tension, but constructions occur in which a bolt may be 
placed in shear. 

The strength of a bolt in shear is about f that of a bolt 
in tension; that is, the shearing strength of wrought iron 
is about -f- the tensile strength. Hence, the diameter of a 
bolt in single shear should be 4/f = 1.1 that of a bolt in 
tension under the same load ; and the diameter of a bolt in 



double shear should be 
under the same load. 



4 



4X2 



= .8 that of a bolt in tension 



1955. Knuckle Joint. — The knuckle joint, Fig. 642, 
is an example of a bolt in shear. Since the bolt is in double 
shear, it need be theoretically only .8 the diameter of the 
rod. The bolt wears, however, and since it should at no 
time be less than . 8 the diameter of the rod, the bolt and 
rod are made equal in diameter. 



MACHINE DESIGN. 



1245 



The other proportions are in terms of the diameter of the 
bolt. All dimensions are in inches. 




Fig. 642. 



Other examples of bolts in shear may be seen in pin- 
connected iron bridges. 



EXAMPLES FOR PRACTICE. 

1. Calculate the diameter of a wrought-iron bolt which is to sus- 
tain a varying load of 2,300 pounds. Ans. I in. 

2. What steady load may be safely sustained by 5 bolts 1^" in 
diameter ? Ans. 13.86 tons. 

3. A screw with square threads transmits motion to a load of 1,500 
pounds. Calculate the diameter of the screw, the number of threads 
per inch, and the necessary number of threads in the nut. 



J 



Diameter = | in. 



Ans. •< Threads per inch = 6. 
( Threads in nut = 10. 



KEYS. 

1956. Keys are iron or steel wedges used to secure 
wheels, cranks, or pulleys to shafts. It is the duty of the 
key to prevent the relative rotation of the pieces connected ; 
if, for example, the pieces in question are a pulley and 
shaft, the function of the key is to prevent the pulley from 
turning on the shaft. Generally, the key will also prevent 
a wheel or pulley from moving lengthwise along the shaft. 



1241) 



MACHINE DESIGN. 



FORMS OF KEYS. 

1957. The concave key is shown in Fig. 643. The 
key is hollowed out to fit the shaft, and holds by friction 
alone ; hence, it is suitable only for light work. 

Fig. 644 shows a shaft with a flat key. A fiat surface 
is cut on the shaft to receive the key, which is, consequently, 
more effective than the concave key. 

1958. The sunk key. Fig. 645, is much more effective 
than either of the above mentioned, since it is impossible 
for the pulley to slip on the shaft without shearing off the 
key. 

A slot called the key-nvay is cut lengthwise in the shaft, 




a 

Fig. 643. 




and another one is cut in the hub of the pulley. The key is 
accurately fitted and driven in. 

Two kinds of sunk keys are in common use: 1. The 




Fig. 646. 



rectangular keys of the form shown in Fig. 646, which are 
used for fastening cranks, gear-wheels, etc. 2. Square keys, 



MACHINE DESIGN. 



1247 



Fig. 645, which are used on machine tools, and, in general, 
on work requiring accurate fitting. 

The keys of the former class are driven in tightly and 
usually fit at the top and bottom as well as at the sides ; 
they are unsuited for parts which require nice fitting, be- 
cause they are liable to spring the parts out of true. The 
square key, on the other hand, does not fit tightly at the 
top and bottom, but rather at the sides. 



1959. Large wheels or pulleys may be fastened to the 
shaft by using two, three, or four keys. Fig. 647 shows a 
method of keying a piece to a square shaft. When several 
keys are thus used the wheel or pulley may be centered on 





Fig. 647. 



Fig. 648. 



the shaft by means of the keys. If a pulley is accidentally 
bored a little too large for the shaft, it may be prevented 
from rocking by using both a sunk key and a flat key, as 
shown in Fig. 648. 

The flat key is placed at a distance around the shaft of 
about 90° from the sunk key, and the pulley is thus made to 
bear on the shaft at 
three points. 

When a key cannot 
be conveniently driven 
out from the small end, 
it is necessary to make it with a gib head a^ as shown in 
Fig. 649. The head forms a shoulder to drive against. 




a 




Fig. 649. 



1348 



MACHINE DESIGN. 



1900- Sliding, or feather, keys are used where it is 
desired to prevent a piece from rotating on a shaft, and, at 
the same time, allow it to slide lengthwise. The key is 




Fig. 650. 



Fig. 651- 



usually fastened to the piece, and is free to slide in the 
keyway of the shaft, though the operation is sometimes 
reversed, and the key is fastened to the shaft. Various 
methods of fixing the key to the wheel or pulley are shown 
in Figs. 650, 651, and 652. In Fig. 652, the key is dove- 
tailed in section, and driven tightly into the hub in. 

When the hub of a wheel in which there is a feather abuts 
against a collar or a bearing it is evident that the feather 
must not project, and in such a case the feather key. Fig. 
652, or the flush feather key. Fig. 651, may be used ; other- 
wise, the key may have gib heads, as shown in Fig. 650. 



1961. Round, or pin, keys may be used when the 
piece is shrunk on to the shaft, as for example, a small 
crank, as shown in Fig. 653. A hole is drilled partly in the 
shatt and partly in the crank, and a round pin is driven in 
the hole, as shown in the figure. 



MACHINE DESIGN. 



1249 



1962. To facilitate the driving in and the removal of 
keys they are usually tapered. The taper varies from -^^ to 
or -j-iirj the smaller tapers being used on the most 



r3"£r 





Fig. 652. 



Fig. G53. 



accurate work. By a taper of -^ is meant that the decrease 

in thickness is -^ the length of the key. Square keys and 

feather keys do not require a taper. 

Note. — To prevent any misunderstanding, the word taper, when 
used in this subject, will mean the gradual diminution in size of a slender 
object. Thus, should it be stated that a certain conical piece has 
a taper of 2 inches per foot, it would be meant that were the conical 
piece one foot long the diameter at one end would be 2 inches larger 
than at the other. 



STRENGTH AND PROPORTIONS OF KEYS. 

1963. A sunk key is subjected to two kinds of stresses. 
The twisting of the piece on the shaft tends to shear the 
key, and also to crush it by compression. 
Let b = width of key in inches ; 

/ = thickness of key in inches; 
/ = length of key in inches ; 
Sg = sa/e shearing stress allowable in pounds per 

square inch; 
Sg = safe crushing stress allowable in pounds per 
square inch; 



Z>. 0. LlL—ld 



1250 MACHINE DESIGN. 

P= force in pounds acting at rim of wheel or pulley; 
R = radius of wheel or pulley in inches ; 
^= diameter of shaft in inches. 

The shearing area of the key is d /; hence, the safe resist- 
ance of the key to shearing is b I S^. Taking moments 
about the center of the shaft, we have 

blS,X^d=PxR, 
^/=-^. {a) 

Suppose the key to be half bedded in the shaft, the crush- 
ing area is f / /, the resistance to crushing \ t I S^. If the 
key is designed to be equally strong against shearing and 
crushing, the shearing resistance must equal the crushing 
resistance, or 

blS, = \tlS,, 

or b = kt% (b) 

If now we assume the crushing strength of the material 

to be double the shearing strength, -^ = 2 ; and we obtain 

b=^t. So is really not double 5^ , but on account of the 
friction between the key and shaft, there is little danger 
of crushing, and a small factor of safety may be used. 
In any case, b is not to be less than /, and for practical 
reasons it is generally made greater. 

For shearing, we may use a factor of safety of about 10, 
giving a safe shearing stress 5^ of 5,000 1b. per sq. in. for 
wrought iron, and 7,000 lb. for steel. Then equation {a) 
above becomes 






(230.) 



Instead of the twisting moment P R oi formula 230, 
it may be more convenient to use the horsepower transmitted 
by the shaft, and its number of revolutions. 



MACHINE DESIGN. 1251 

Let N = number of revolutions per minute ; 
//= horsepower. 

Then, a point on the circumference of the wheel or pulley 
moves 2 TT ic iv inches per mmute, or — — — • feet per mmute. 

Hence, if a force P constantly acts at the circumference 
of the wheel, the work done per minute is — -^r — X P 

I/O 

foot-pounds. 

o jj. J? Af P 
Therefore, — '- 33,000 = //", the horsepower. 

1^2<_|»^= Pie = 63,025 f. (231.) 

Formula 231 will be frequently used hereafter, and 
should be carefully studied. 

Substituting the value oi P R from formula 231, in 
formula 230, we have 



1= J ^j. for wrought iron. 
afv 

ifl= J ,^ for steel. 



(232.) 



Formulas 230 and 232 may be used in calculating the 
sizes of keys for large work. For small shafts the sizes 
given by 230 and 232 are much smaller than are used in 
actual practice. 

1 964. Designers usually adopt some standard ratio be- 
tween the depth and width of the key, the ratio varying from 
•J- to |-. We shall adopt the ratio f ; that is, / = f d. 

Example. — The maximum pressure on an engine crank-pin is 12,500 
pounds, and the length of crank is 10 inches. Suppose the diameter 
of the shaft to be 5 inches, and the length of the key the same. What 
should be the dimensions of a wrought-iron key to hold the crank to 
the shaft ? The crank is not to be shrunk on the shaft. 

Solution. — Using formula 230, 

u_ PR _ 12,500X10 _g. f.^^ 

2,500^/ 2,500 X 5 X 5 ~ 
/ = |^ = f X2 = li", say li^'. Ans. 



1252 



MACHINE DESIGN. 



1 965. In common designing, the sizes of keys are deter- 
mined by empirical formulas, which give an excess of 
strength. For an ordinary sunk key, the following propor- 
tions may be adopted: 

b 
t. 



id. 



= ^d. \ 



(233.) 



Using formula 233 in the example of the crank-shaft 
above, 

b = ld=iX^ = lV' 
t = ^d=\=\^\ nearly. 

The key is sunk for ^ its depth in the shaft. 

The following empirical formulas give good results; 





d 




1^ ^ 


b^ 


''Y 


+ 


4' 




d 




Z" ' 


t — 


12 


+ 


16' 



For driven pulleys. 






When d is less than 1^^ in. , 



b=: 



d_ 
3' 

£ 
5" 



(234.) 



(235.) 



(236.) 



Formulas 233 and 235 give nearly the same results ; 

24- h" 
thus, for a 2^-inch shaft, 233 gives ^ = --p = -— , and t = 

f^ nearly. 235 gives ^ = .66 in. = \\\ and / = .41" = -f-^" . 
Unless otherwise stated, the student may use formula 233 
in solving his problems. 

For sliding feather keys, the following formulas give the 
proportions used in ordinary practice : 



8 ^ 16 
16 ^16 



(237.) 



MACHINE DESIGN. 1253 

1966. In some instances pulleys may be keyed to a 
large shaft and yet transmit a small amount of power. In 
such cases the dimensions of a key, if based upon the actual 
diameter of the shaft, would be much too large, and the 
proportions should be based upon the diameter of a shaft 
which would be necessary to transmit the power of the pulley 
in question, and no more. Letting H represent the horse- 
power transmitted by the pulley, and N represent the 
revolutions per minute of the shaft, we may take 




^=5y-^, (238.) 

and use this value of d in formula 236. 

Example. — A pulley transmitting 2 horsepower is keyed to a shaft 
4 inches in diameter making 120 revolutions per minute. Determine 
the dimensions of the key. 

Solution. — The diameter of a shaft to transmit 2 horsepower is 
d= ^i/tL = 5 i/— = 1.28". Ans. 

y jv r 120 



From formula 236, 

^_ g- g _^. Ans. /_ ^ _ ^ 



EXAMPLES FOR PRACTICE. 

1. Calculate the dimensions of an ordinary sunk key for a shaft 
3| inches in diameter. Ans. y\" X i". 

2. Calculate the dimensions of a feather key for a shaft 2f inches 
in diameter. Ans. if" X -^Z- 

3. A wrought-iron key is used to fasten a fly-wheel on a 6-inch 
shaft. If the maximum pressure on the crank-pin is 15,000 1b., and 
the crank radius is 16 inches, what should be the dimensions of the 
key, its length being 8 inches ? Ans. 2" X IH"- 

4. A pulley transmits 3^ horsepower, and is keyed to a shaft 
5 inches in diameter, making 150 revolutions per minute. Calculate 
the dimensions of the key. Ans. Y X tV» 

COTTERS. 

1967. A cotter is an iron or steel bar which is driven 
through one or both of two pieces to be connected, and holds 
them together by its resistance to shearing at two transverse 



1254 



MACHINE DESIGN. 



cross-sections. An example of a cotter was shown in Fig. 
617. The cotter C passes through the foundation bolt, and 
is subjected to a shearing stress along the dotted lines. 




Fig. 654. 

A simple form of a cotter is shown in Fig. 654. The 
cotter passes through the rod only, and acts when the rod is 

in tension. The enlarge- 
ment, or collar a^ on the 
rod prevents any down- 
ward movement, and, 
therefore, resists thrust. 

Fig. 655 shows a cotter 
with gib ends. Since in 
this case the rod is not 
provided with a collar, 
this arrangement will re- 
sist tension only. In the 
arrangement shown in 

Fig. 656, the cotter is 
Fig. 655. ,..,,.' , 

divided into two parts, the 

one with hooked ends being called the gib, and the other the 

cotter. 




MACHINE DESIGN. 



1255 



In this construction the rod should be placed in tension 
only. If it is necessary to provide for thrust the end of the 
rod should be tapered as shown in Fig. 657. 





Fig. 656. 



Fig. 658 shows an arrangement in which a rod is cottered 
into a socket. As shown in Figs. 656 and 658, the cotter is 

long and tapered; it serves, 
therefore, as a means of adjust- 
ing the length of the connected 
pieces. By driving the cotter 
farther in, the total length of 
the two pieces is lessened, and 
vice versa. 

A cotter may be used to con- 
nect two straps in and n to a 
rod / shown in Fig. 659. When 
driven down the friction between 
the cotter and lower strap must 
cause the latter to open out as 
shown by the dotted lines. Hence, it is desirable in such a 
case to use a cotter combined with a gib, as shown in Fig. 
660, or with two gibs as shown in Fig. 661. 

The gibs serve to keep the straps from spreading. In 
Fig, 662 the side a b oi the gib and c d oi the cotter 




Fig. 657. 



1256 



MACHINE DESIGN. 



are parallel to each other, and perpendicular to the axis 
of the rod, and the taper comes between the gibs and 
cotter. 





Fig. 658. 



1968. Strength and Proportions of Cotters. — In 

designing a cotter connection the following points must be 
taken into account: 




Fig. 659, 



Referring to the illustration of the cotter, Fig. 654 — 



MACHINE DESIGN. 



1257 




1. The cross-section b t oi 
the cotter must be sufficient 
to withstand the shearing 
force. 

2. The thickness t must be 
great enough to provide 
against failure by crushing. 

3. The two diameters d^ 
and d should be so designed 
that the rod is of uniform j^fti^^jAji 
strength throughout. fig. eei. 

Let P = force in pounds exerted on the rod ; 

S^ = safe shearing strength of cotter in pounds per 

square inch; 
Sc = safe compressive strength of cotter or rod in 

pounds per square inch; 
S^ = safe tensile strength of rod in pounds per 

square inch. 

The various diameters and other dimensions are indicated 
on the figures. 

Consider the arrangement shown in either Fig. 656 or 
Fig. 657 and conceive a section taken through the cotter 

hole. The net area of the rod is { — d'^ — dt\ very nearly; 

the shearing area of the cotter is 2 ^ /; the area of the cotter 
subject to crushing is dt^ very nearly; the area of the socket 

subject to tension is — (/?' — d"^) — {D — d)t\ finally, the 

area of the smaller part of the rod is ^T^d^. 

Hence, P=(^d''- 



dt\s,. 



P=2.5tS,. 
P=dfS., 



P=[l{n^-d^)^{i^-d) t^s,. 



P^^d^S,. 



(a) 

(d) 



1^58 MACHINE DESIGN. 

Suppose both rod and cotter are made of the same 

S 4 
material, either wrought iron or steel, and take -^ = — , as 

was done before in Art. 1954. 

5 
Experience shows that S^ may be double S^ ; that is, -^ = 2. 

Hence, -^ = - = 2^. 
Now, combining equations (d) and (^), 

Combining equations (a) and (^), we have 
/^^^ _ dt\ = dt^ = 2 dt; 

whence, 2> dt— -jd^^ and t = j^d=z .26 d= say ^d. 

Combining equations {a) and (d), and taking t — -^d, it 



will be found that 






TT ^ 

Combining equations (a) and (^), and taking t = — -, 

—d^ = —d^ — dt = —d'^ — -xd^; whence, 
4^4 4 12 

d^' = %d\ and d^ = .S16d. 

It was shown above that to have the same tensile strength 
as the rod, the diameter D of the socket or boss should 
be 1.3 d. 

To prevent failure from crushing, however, the bearing 
surface of the socket should equal that of the rod, or 

{B — d) t = d t. Hence, D = 2 d. 

The diameter d^ of the collar, Fig. 658, should be such 
that the bearing surface of the collar is at least equal to 
that of the cotter in the rod. Hence, 



MACHINE DESIGN. 1259 



li^'-^') 



dt = ^cl^.. 



whence, - d^ =■ - ^', and d^ = d\/^ — 1.15 d. 
Collecting the above results, we have 

b =lid; 

t =id; 

d^ = .SUd; 

I)=2d; 

d^=1.15d; 

h = c = ^d to l^d. 



(239.) 



1969. If a steel cotter be used in a wrought-iron rod, 
d may be made equal to d, the other dimensions remaining 
the same as above. 

For a cotter of the form shown in Fig. 660 or 661 it is 
good practice to make / = ^ ^, and d t = ^ X sectional area 
of strap. 

The width d is the same whether a single cotter, a gib 
and cotter, or two gibs and cotter are used. The other 
proportions are shown on the figures. 

Example. — Suppose in Fig. 661 the etrap is I" X 3^"; find the thick- 
ness of the cotter, and combined width of gibs and cotter. 

245 
Solution.— (5 ^ = ^XiX^= -wr^- Making / = J <5, as stated above, 



whence, d = i/ ^^^ = S.QV = 3^f". Ans. 



^ 245 
16 
/ = i <^ = 1 inch, nearly. Ans. 

The width of cotter is, then, f d^ or l^y ^^^ the width of 
each gib is ^ d, or IjV* 

1970. Taper of Cotters. — The greatest allowable 
taper that a cotter may have without danger of slacking 
back is about ^. 

Usually, the taper is -^^ to ^^g- when the cotter is not 
secured. If fastened by a set-screw or bolt and nut the 



1260 



MACHINE DESIGN. 



cotter may have a taper of -J- or ^. The taper is found by 
dividing the increase in width by the length of cotter. 
Thus, if a cotter is 2 inches wide at one end, 2^ inches 
wide at the other, and 12 inches in length, the taper is 
m - 2) -M2 = ^. 

1971. Locking arrangements for cotters are 

shown in Figs. 662 and 663. In Fig. 662 the cotter is held 



aX 




H 



u 



~J 



] 




Fig. 662. 

by a set-screw, the point of which fits into a groove cut into 
the cotter. The diameter of the set-screw may be ^ ^ + i"' 
In the arrangement shown in Fig. 663 the end of the cotter 

is a screw, and the 
cotter is secured by 
a nut on an extra 
seat. This method 
is used where the 
cotter has an exces- 
sive taper. 

1972. A split 

pin is a form of 
cotter which is used 
not to firmly con- 
nect two pieces, but to prevent them from separating en- 
tirely. Small split pins are of the form shown in Fig. 664. 
When large pins are used for this purpose they are solid and 
tapered. 




Fig. 663. 



Fig. 664. 



MACHINE DESIGN. 



1261 



r^^l^^in. 


; <5 = U in. ; 


^1 = 1 in. ; 


t = \\m.; 


di - 1| in. ; 


D = ^in.; 


h = 1 in. 





EXAMPLES FOR PRACTICE. 

1. Find the dimensions of a rod and socket of the form shown in 
Fig. 658, assuming St = 6,000 pounds. The load or pull on the rod 
is 4,600 pounds. 

Ans. 



2. A cotter of the form shown in Fig. 655, resists a pull of 3,200 
pounds. Find the necessary breadth and thickness on the assumption 
that Ss is 4,000 pounds, and that the thickness is one-fourth the breadth. 

I f = j% m. 

3. A cotter and two gibs connect two straps to a rod, as shown in 
Fig. 661. Supposing the pull on the rod to be 9,000 pounds, and taking 
Ss = 5,400 for steel, find the dimensions of cotter and gibs. 

( / = if in. 
Ans. •< Width of cotter = ^^ in. 
( Width of gibs = ^^ in. 

4. In example 3, (a) what should be the «<?/ section of the strap to 
be equal in strength to the cotter ? Assuming the thickness of strap 
to be i the width, {d) what would be its actual dimensions ? 

Ans. I (^) -^^ ^q- ^^• 



(<^) 2i in. X U in. 

5. Calculate the dimensions of a steel cotter which fastens a wrought- 
iron rod 2f in. in diameter. Ans. 2f in. X if in. 

6. A cotter is If inches wide in the middle and tapers on each side. 
If the cotter is 18 inches long, what is its width at each end ? Assume 
that the taper is i inch to the foot. Ans. 2^ in. and 1| in. 



ROTATIIVG PIECES. 



JOURNALS. 

1 973. Journals are the cylindrical portions of rota- 
ting pieces which turn within bearings and form the supports. 
Journals which are situated at or near the end of a shaft, 
axle, or other rotating piece, 
are termed end journals. 
Any journal situated between 
two end journals is called a 
neck Journal. 

The ordinary form of an fig. 665. 

end journal is shown in Fig. 665. It consists simply of a 



n-^fi 


P 

\f 






1 
d 



1262 MACHINE DESIGN. 

cylinder with collars at each end to prevent end play in the 
bearing. 

The length of the step, seat, or bearing on which the 
journal rests, is, however, often made slightly shorter than 
the journal, permitting a slight motion lengthwise, and 
securing uniform wear. 



SIZE AND PROPORTIONS OF END JOURNALS. 

1974. The chief element in the design of a journal 
moving slowly or intermittently is strength. When journals 
run constantly at considerable velocity, strength is not so 
important a consideration as durability and freedom from 
liability to heat. 

The dimensions of an end journal to give sufficient 
strength may be calculated by considering the journal as a 
cantilever uniformly loaded. 

Let / = length of journal in inches; 
d = diameter of journal in inches; 
P = total load on journal in pounds; 
Sj> = safe stress of material in flexure. 

Then, from the Table of Bending Moments, the bending 
moment is — ^— = -— , and the resistmg moment is -^ — , 

<j Ai C 

which for a circular section is (see Table of Moments of 
Inertia), 

ltd' 
^ 64 _ ^ 7r^» 
^' \d -^^"32"- 

Hence, — = ^/-^, (p) 

Formula 240 gives the diameter of the journal when 

the ratio -j is assumed. 
a 



or 



MACHINE DESIGN. 1263 

Example. — Find the diameter and length of a wrought-iron journal 
on which there is a load of 1,200 pounds. Assume 5/ = 8,500 pounds 

and -J — 1.4 
a 

Solution. — Using formula 240, 



^=2.26|/^X ^ = 2.26j/i|5?X 1.4=1", nearly. Ans. 
/= 1.4 ^=1.4". Ans. 

1975. The bearing surface, or projected area, of 

a journal is the length multiplied by the diameter; that is, 
it is the area of the projection of the journal on a plane. 
The total load on the journal divided by the projected area 
gives the pressure per square inch of projected area — a 
quantity which will be denoted by /. 

p 
Hence, p = j-j, or P=p Id. (J?) 

In order that the journal may not heat, the pressure p 
must not exceed a certain limit determined by experience. 
When this pressure is too great, the oil used to lubricate the 
journal is squeezed out, and the journal heats rapidly. 

From equation (^), F— ' . 

From equation (U)^ P = pld. 
TZ S d^ 

Hence, t^, =pld^ or izSfd"^ =1Q p l^\ and 

J.0 If 



d V 16/ 



W 



Substituting this value of --j in 240, we obtain, after 
a slight reduction, 

d= J -^^4- = 1- 5 J-^ (241.) 

p 

From equation (b), I = -j-. (242.) 



1264 MACHINE DESIGN. 

Formula 241 may be used to compute the diameter of a 
journal when the pressure / per square inch of projected 
area is fixed. The length may then be obtained from 
formula 242. 

Example. — Compute the length and diameter of a steel journal 
sustaining a load of 13,000 pounds. The safe stress 5/ is 14,000 pounds, 
and the pressure per square inch of projected area is not to exceed 750 
pounds. 

Solution. — Using formula 241, 



. / P , / 13,000 



^='-'^ T^^'-''^' t- 14,000 X 750 = ^-^^'> ^^^^^'' ^- 



P ^ 13,000 
^/~3ix750 



Hence, / = -^j-^ = ^ ^ l^ ^^.^ = 5j\". Ans. 



1976. The pressure/ per square inch of projected area 
may be taken at from 400 to 800 pounds, when the journal 
runs constantly at a speed under 150 revolutions per minute. 
For journals which run slowly or intermittently, p may be 
much greater, while .for journals running faster than 150 
revolutions per minute, the pressure / should vary inversely 
as the number of revolutions. That is, letting N = number 

of revolutions per minute, / = — , where <3: is a constant. 

Another consideration affecting the allowable pressure / 
is the direction of the load. In some journals the load acts 
only in one direction, generally downwards ; in others, as, 
for example, crank-pins and cross-head pins, the direction 
of the load changes at every revolution. In the latter case, 
the pressure / may be twice as great as in the former, 
because the change in the direction of the load permits a 
more perfect lubrication of the bearing. When, however, 
the direction of the load is variable, the safe stress Sf 
must be taken smaller than when the direction is con- 
stant. 

1977. The following values of / for different kinds of 
journals are taken from Unwin: 



MACHINE DESIGN. 1265 

TABLE 45. 



PRESSURE ON BEARINGS AND SLIDES., 

Pressure per 

*^. , ^ -r * ^ . Sq. In. of 

Kind of Journal Bearing. Projected 

Area, p. 

Bearings on which the load is intermittent and 
the speed slow, such as" crank-pins of shearing 

machines 3,000 lb. 

Cross-head neck journals , 1,200 lb. 

Crank-pins of large, slow-speed engines. . . .800 to 900 lb. 

Crank-pins of marine engines, usually 400 to 500 lb. 

Main crank-shaft bearings — marine engines (slow) 600 lb. 

Main crank-shaft bearings — marine engines (fast) 400 lb. 

Locomotive driving axle journal 180 to 350 lb. 

Railway journals 200 lb. 

Fly-wheel shaft journals 150 to 250 lb. 

Small engine crank-pins 150 to 200 lb. 

Slipper slide blocks, marine engines 100 lb. 

Stationary engine slide blocks 25 to 125 lb. 

Stationary engine slide blocks, usually 30 to 60 lb. 

Propeller thrust bearings 50 to 70 lb. 

Shaft in cast-iron steps or seats 15 lb. 

For journals not given above, the value of / must be deter- 
mined by the judgment of the designer. The value adopted 
should seldom or never exceed 750 pounds. For journals 
running faster than about 150 revolutions per minute, / 
should vary inversely as the number of revolutions. For 
example, if a given journal is allowed a bearing pressure of 
300 pounds at 150 revolutions, it should only be allowed a 

bearing pressure of • — — -- — = 180 pounds if it is required 

to run at 250 revolutions. 

1978. The permissible working stress Sf may be taken 
as follows: 

2>. 0. 111.— 20 



1266 MACHINE DESIGN. 

TABLE 46. 

Direction of Load Constant. Direction of Load Variable. 

Steel Sj,= 14,000 Steel 5^= 12,000 

Wrought iron ...Sf= 8, 500 Wrought iron ...Sf= 7, 000 
Cast iron Sf= 4,000 Cast iron Sf= 3,000 



NECK JOURNALS. 

1 979. A neck journal similar to an engine cross-head 

pin may be considered as a beam supported at both ends 

and uniformly loaded. Consequently, the bending moment 

wP PI 
is — ^— = ---. (See Table of Bending Moments and Defiec- 

o o 

tions. ) 

Hence, "o" = -"%o — » (^^^ (^)> -^^^- 1974), 



and^=7if^x4=1.137|x-^. (243.) 

Formula 243 may be used to calculate a neck journal 
when —J is given or assumed. 

But, P=pdl, (see (<5), Art. 1975). 



Hence Pi^tAL-^ll^ or ^ - i/^ id\ 
Hence, — --^---^^, or -^-y—. (^) 

Substituting this value of -j in formula 243, we obtain 



^=1.06|/-^. (244.) 

¥tomP=pdl, ^=Jd' (^45.) 

The same values of Sf and / may be taken as for end 
journals. 

It will be seen by comparing formulas 241 and 244 
that for the same load the neck journal need be but about 
J the diameter of the end journal; and a comparison of 



MACHINE DESIGN. 1267 

equations (c) and (d) shows that the ratio -7 is double in a 

neck journal what it would be in an end journal. 

Note.— In using formulas 241 and 244 it will be found most con- 
venient to use logarithms. Thus, applying logarithms to formula 241 
it becomes log ^ = log 1.5 + | [log P — i (log / + log S/)]. 

Example. — Find the length and diameter of a wrought-iron neck 
journal, the load being 9,600 pounds, variable in direction. Allow a 
bearing pressure of 600 lb. per square inch. 

Solution. — Using formula 244, 



'=lMy—^=r=1.06^- 



9,600 

or 



Vj^ ^ |/'600 X 7,000' 

log d = log 1.06 + i [log F-i (log/ + log Sf)] = 

.03531 + i [3.98227 - -^(2. 77815 + 3.84510)] = .36063; 

whence, d= 2.295" = 2^', or nearly 2^^^". Ans. 

, P 9,600 „„ , 

/ = ^ , = TJTTTT — TT-r- = 7 , nearly. Ans. 
p d 600 X 2j\ 

1980. It may sometimes be desirable to make the 
length of a journal greater than the calculated value, simply 
as a matter of taste. To give the journal the same strength, 
its diameter must be increased in the proportion given by 
the following formula : 



't-m- 



(246.) 



where / and d are the original and /^ and d^ the new lengths 
and diameters of the journal, respectively. 

For example, suppose a wrought-iron end journal to be 

subjected to a load of 350 pounds. Assuming -r = 1.4, the 

dimensions of the journal will be 

■ ^= 2.36 V^ X ^ = 2-^6 ^^^ X 1.4 = .5426'. 

/= lAd= 1.4 X .5426 = .7596^ 

Say the diameter is ■^" and the length |^ 
Though a journal of these dimensions would be suffi- 
ciently stiff and durable for the load, it is rather small to 



1268 MACHINE DESIGN. 

look well. Suppose the length is made double the calcu- 
lated value ; that is, /^ = 2 /. Then, from formula 246, 

d^ = ^-v/4 = .5426-V^= .6837" = \^\ nearly. 

/, = 2/=2x| = 4. 



FRICTION OF JOURNALS. 

1981. The work expended in overcoming the friction 
of a journal is 

TT d 

W= fPN—^ foot-pounds per minute; 

where /*= coefficient of friction, 
and N = revolutions per minute. 

It is therefore apparent that with the same load and speed, 
the work expended against friction is directly proportional 
to the diameter of the journal. Consequently, when a jour- 
nal runs under a steady, uniform load, it is preferable to 
make the diameter as small as possible, consistent with 
strength, and obtain the desired projected area by adding to 
the length of the journal. 

For example, take two journals, one 2 inches in diameter 

and 6 inches long, the other 4 inches in diameter and 3 

inches long. They each have the same projected area, 12 

inches. The latter journal, however, requires double the 

work to overcome friction that the former requires, and, 

besides, contains twice as much material. Hence, the 

2" X Q" journal is preferable for a steady load. On the 

4' X 6 
other hand, the 4''' X 3'' journal is -^ =16 times as strong 

as the other, and would be preferred in situations where the 
load is variable, and the journal is liable to shocks, as in the 
case of crank-pins of high-speed engines. 

1982. The height of the journal collars, Fig. 665, may 
be 



MACHINE DESIGN. 



1269 



The width of- the outer collar is 1^ e. It is good practice 
to turn the journal with a fillet in the corner, as the shaft or 
axle is more liable to crack and fracture if the shoulder is 
turned with a square corner. 

The fillet may be a circular arc drawn with a radius equal 
e 



to 



2* 



PIVOTS AND COLLAR JOURNALS. 

1983. A pivot journal is shown in Fig. 666. It 
differs from an ordinary journal in that the 
direction of pressure is parallel to the axis 
of the shaft instead of perpendicular to it. 
The bearing area is, therefore, the area of 
the end of the pivot, \ tt <^^ 

The diameter of a flat pivot may, there- 
fore, be found at once by assuming a 
value for the pressure per square inch of 
projected area and solving for the area. The following for- 
mulas give good results for ordinary cases: 

Let /*= load on pivot ; 

d =^ diameter of pivot; 

iV*= revolutions per minute of pivot. 

TABLE 47. 




Fig. 666. 



Wrought-Iron 

or Steel Pivot 

on Gun-Metal 

Bearing. 



Cast-Iron Pivot 

on Gun-Metal 

Bearing, 



Iron or Steel on Lignum- 

vitae Bearing, 
Moistened with Water. 



Case I — Pivot turn- 
ing very slowly or 
intermittently . . . 

Case II — Revolu- 
tions per minute 
less than 150 

Case III — Revolu- 
tions per minute 
more than 150. , . 



d=. 035 |/^ 



^=.05 \/T. 



d=.05i/F. 



d^m J/P. 



(248.) 



^=.004|/;^A^. 



^=.035i/p: (249.) 

d=MhifP. (250.) 



1270 



MACHINE DESIGN. 



Example. — What should be the diameter of a steel pivot on gun- 
metal bearings, the load being 300 lb. , and 
the number of revolutions 320 per minute ? 
Solution. — 
d= .004 \/PN= .004 |/200 X 320 = 1", 
nearly. Ans. 




Fig. 667. 



1 984. In a collar journal the 

direction of the load is parallel to the 
axis of the shaft, but the bearing area 
is formed by a collar raised on the 

shaft. A journal with one collar is shown in Fig. 667, and 

one with several collars in Fig, 668. 

Letting d^ and d represent the diameters of collars and 

shaft, respectively; b^ 

the number of collars ; 

/), the pressure per 

square inch of pro- 
jected area, and /*, 

the total load or 

thrust, we have 



^7:{d^'-d') b p=P. 

The value of/ is fig. 668. 

usually taken at 60 lb. per square inch for the thrust bear- 
ings of propeller shafts, and this value should not be much 
exceeded in any case. 

p 

Then, d^^ - d'' - --—7, 




or d. 



; = / 



d'-V 



F 



15 TT^' 



(251.) 



The number of collars b depends upon the judgment of 
the designer. The larger the number the smaller is the 
diameter, and the less will be the wear and work of friction. 
On the other hand, when many collars are used, there is 
danger of bringing all the thrust on one or two. 

In Fig. 668 the dimension e ^^ \ (d^ — d). Usually the 
thickness t of the collars is = .8 ^, and the width s of the 
space is equal to the thickness /', unless the annular encircling 



MACHINE DESIGN. 1271 

rings k are hollow for water circulation, or unless they are 
lined with white metal. 

In the latter case, .$■ = 2 to 2|- ^. 



EXAMPLES FOR PRACTICE. 

1. Find the proportions of a cast-iron end journal, turning slowly 
under a steady load of 15,000 lb., assuming the length equal to the 
diameter. Ans. 41" X 4f ". 

2. Find the proportions of a wrought-iron end journal, turning 
under a load of 8,600 lb., assuming the safe pressure per square inch of 
projected area to be 650 lb. The direction of the load is variable. 

Ans. 3"x4yV- 

3. Find the dimensions of a steel neck journal, working under a load 
of 13,000 lb., direction variable, the allowable bearing pressure to be 
1,200 lb. per sq. in. Ans. 1|" x S^V- 

4. Find the minimum dimensions of a wrought-iron end journal, 
the required conditions being that the bearing pressure shall not ex- 
ceed 750 pounds, and the length shall be one and one-fourth times the 
diameter. The load on the journal is 9,600 pounds, and variable in 
direction. Ans. 3^" X 4". 

5. Find the diameter of a wrought-iron pivot running at a speed of 
80 revolutions per minute and bearing a load of 800 pounds, gun- 
metal bearing. Ans. 1^ in. 

6. Assuming 5 thrust collars on a shaft 10 inches in diameter sub- 
jected to an end thrust of 20,000 pounds, find the diameter and thick- 



ness of collars. . (^1 = 13.6". 

Ans. i . 



A'; 



SHAFTS. 

1 985. Shafts may be divided into three classes, accord- 
ing to the kind of stress to which they are subjected: 

(1) Shafts subjected chiefly to torsion or twisting, as, for 
example, line shafting in mills and shops, and, in general, 
shafts used to transmit power. 

(2) Shafts subjected chiefly to bending action, such as 
the axles of gears, etc. 

(3) Shafts subjected to both twisting and bending, such 
as engine crank-shafts. 

LINE SHAFTING. 

1986. Lrine shafting is a term applied to the long 
and continuous lines of shafting used in mills, factories, and 
shops for the distribution of power. The shaft is principally 



1272 



MACHINE DESIGN. 



strained by torsion, but there is always in addition a bend- 
ing action due to the weight of the shaft itself, and the 
pulleys carried by it, and also due to the tension of belting 
or the thrust of gearing. 

In calculating the diameter of a shaft for a given twisting 
moment, two things must be considered : 3 Strength. 2. 
Stiffness. Very large shafts or short shafts need be calcu- 
lated for strength only; but in long lines of shafting of 
small diameter attention must be paid to stiffness and 
rigidity. 

The twisting moment is P R, where P is the twisting 
force and R the distance between the line of action of the 
force and the center of the shaft. In case of a belted pulley, 
shown in Fig. 669, the force Pis the difference in the ten- 




FlG. 669. 



sions of the tight and slack sides of the belt ; the distance R 
is the radius of the pulley. Or, again, R may be the radius 
of a gear attached to the shaft in question. 

It is usually more convenient to express the twisting 
moment in terms of the horsepower transmitted by the 
shaft, and number of revolutions. 

TT 

By formula 231, PR = 63,025 -^. The resistance of a 
shaft to twisting is given by a formula similar to formula 



IIQ, M=S,^, 



Art. 1398. 



MACHINE DESIGN. 1273 

Let Sg = the safe shearing strength of shaft ; 

y = polar moment of inertia of cross-section of shaft 

about neutral axis ; 
c = distance from neutral axis to outermost fiber. 

, S / 
Then, the moment of resistance to twisting is — ^ — . 

Where a plane section revolves, or is conceived to revolve, 
about a line lying in the plane of the section, as, for 
example, a circular section revolving about its diameter, 
the moment of inertia obtained with reference to this axis 
is termed the rectangular moment of inertia, and is 
nearly always denoted in text-books by /. When, however, 
the section revolves about an axis perpendicular to the 
plane of section, as, for example, a buzz saw or fly-wheel, 
the moment of inertia found with respect to this axis is 
termed the polar moment of inertia, and is nearly al- 
ways denoted in text-books by y. Those moments of inertia 
used in the subject of Strength of Materials were rectan- 
gular moments. For a solid circular section, y"= ^/= 2 X 

-^7- = —^r— ; and for a hollow circular section, 

^ 32 

The reason for using the safe shearing stress S^ in the ex- 

S / . 
pression — ^ — is that its value agrees very closely with the 

value for torsion ; it is quite different from the safe stress 
for flexure. The student may obtain the value of 5^ by 
simply dividing the ultimate shearing stress by the proper 

5 
factor of safety ; i. e. , 5^ = -^. 

Placing the twisting and resisting moments equal to each 
other, we have 



N ~ c d_ 16 

2 



1274 MACHINE DESIGN. 



Whence, ^= /Ii^= 1.72 jA^. (252.) 

_ , ,yi6X 63,035// aoKi/^L /OKr»\ 

Let f^ = ^, and ^^^IZW ^ 




s> 



// 



Then, d= k \/PR=z k^f ^, 



as in formula 124, Art. 1416. 

The values of k and k^ depend upon the value of the safe 
stress 6*3 assumed for the material of the shaft. 

If we take ^^ = 3,400 for cast iron; 

5s = 6,800 for wrought iron; 
5^ = 9,000 for steel, 
we obtain the values of k and k^ given in Table 32, 
Art. 1416. 

1987. Formulas 252 and 253 are general, and may- 
be used for any material ; the proper value of vS^ being left 
to the judgment of the designer. Ordinarily, the values of 
Sg given above are those used in good practice. Under ex- 
ceptional circumstances it may be necessary to use lower 
values, particularly if the shaft is subjected to shocks, or if 
the stress changes suddenly or violently. 

It has been shown that for strength the diameter of a 
shaft is proportional to the cube root of the twisting mo- 
ment. A similar theoretical treatment would show that for 
stiffness the diameter of the shaft is proportional to the 
fourth root of the twisting moment. 

Hence, for shafts of small diameter or of considerable 
length the diameter may be calculated by the formula 

d=zcYPr = c^ y -^ and Table 31, Art. 1415. 

1988. Speed of Line Sbafting. — The speed of a 
shaft is fixed largely by the speed of the driving belt or the 
diameters of the pulleys upon it. In general, machine shop 



MACHINE DESIGN. 



1275 



shafts run about 120 to 150 revolutions per minute ; shafts 
driving woodworking machinery, about 200 to 250 revolu- 
tions per minute; in cotton mills, the practice is to make 
the shaft diameter smaller, and run at a higher speed. Line 
shafts should generally not be less than If inches in 
diameter. 

1 989. Distance Bet^ween Bearings. — This distance 
should not be great enough to permit a deflection of more 
than y^'' per foot of length. Hence, when the shaft is 
heavily loaded with pulleys, the bearing's must be closer than 
when it carries only a few pulleys. 

Below are given the maximum distances between the bear- 
ings of different sizes of continuous shafts which are used 
for the transmission of power: 

TABLE 48. 





Distance Between Bearings 


Diameter 




in Feet 




of Shaft 








in Inches. 


Wrought- 
Shaft. 


Iron 




Steel 
Shaft. 


2 


11 






11.5 


8 


13 






13.75 


4 


15 






15.75 


5 


17 






18.25 . 


6 


19 






20 


7 


21 






22.25 


8 


23 






24 


9 


25 






26 



Pulleys which give out a large amount of power should be 
placed as near a hanger as possible. 



SHAFTS SUBJECTED TO BENDING. 

1990. Under this head are included the axles of large 
water-wheels, gear-wheels, etc., which are loaded trans- 
versely. The axle is not generally strained by twisting, and 



1276 MACHINE DESIGN. 

may be treated as a beam transversely loaded. The mode 
of procedure may, perhaps, be best shown by an example. 

Example. — An axle 12 feet long, between centers of bearings, carries 
a wheel weighing 9 tons ; the wheel is 4 feet from one end of the shaft. 
The axle is of wrought iron. Required, the dimensions. 

Solution. — Let the line m n, Fig. C70, represent temporarily the 
axle in question ; P, the load of 9 tons, and Ri and R^, the reactions. 

Lay off to scale the load P on the vertical through i?i, and, choosing 
a pole 0, draw the rays (91, 3, and ab, and cb parallel to them. 
Then, abc is the bending moment diagram. 6> 2 is drawn parallel to 
a r, and gives the reaction Ri = 1-2 = 3 tons, and R^ = 2-3 = 6 tons. 

The lengths and diameters of the journals may now be calculated. 
For the journal nearest the load, we have, from formula ;241, 



d=1.5i/ ri— y and 
assuming that/ = 700 and 5/= 8,500, 



</=1.5j/- 



12,000 

"f , nearly. 



|/ 8,500 X 700 

7?a 12,000 
^=J^=700x3f=^*'^"^^^y- 



For the other journal, 



, / Ri J 6,000 

</' = 1. 5 i/ li— = 1. 5 i/ , = Sf. 

^ >^/5// ^ ^8,500x700 ^ 

_6^000__ 
^'"•700x2f~^ • 

d^ = d'+%e, where e = ^V ^' + i" = -^625" = f", nearly. 

Hence, d^ = 2f " + f " = 3i". 

Likewise, d^ = d+2{^^d+ i") = 4^". 

The resisting moment of a beam is 5— (see Art. 1398), and since 
the safe stress S remains constant for the same beam, the resisting 

Foi 



moment is proportional to — . For a beam of circular section. 



7 _ 64 _ 7r ^3 
c~ d ~ 33 



MACHINE DESIGN. 



1277 



Hence, since -ho is a constant quantity, the resisting moment at any 

section of a circular shaft is proportional to the cube of the diameter 
at that section. Conversely, the diameter of the shaft at any section 
must be proportional to the cube root of the resisting moment at that 




Fig. 670. 



section, and to have the shaft of equal strength throughout, the diam- 
eter at each section should be proportional to the cube root of the 
bending moment at that section. 






moment b e 



Hence, j — -^ ^ y 

a r moment g h 

By measurement, the moment b e vs, 288 inch-tons, and the moment 
g Ais 15| inch-tons. 

Whence, ^ = i/i| = 2.65; 

or, ^, = 2.65 ^= 2.65 X 3f = 9 inches, nearly. 

The shaft should be bossed at this point to allow for cutting the 
key-way. The depth of the sunk key = ^ di = 1^ in. (See formula 
233.) 

The depth of the key-way is, therefore, f inch, and the diameter of 
the bossed portion 10| inches at least. The diameters d^ and da may 
be easily found by measuring the moments^ ^ and ^ z\ whence, 

d-i = di 4/ |-^, and dz = di47 ~, — 
foe f b e 

In a similar manner may be found the dimensions of an 
axle bearing two or more transverse loads. 



1278 MACHINE DESIGN. 

The axle may or may not be tapered as shown in Fig. 670. 
If it is not tapered, its diameter must be calculated from 
the maximum bending moment {b e, in this case). A 
straight shaft is easier to construct, but the end journals 
must be, of course, larger, and there is considerable loss by 
friction on that account. 



SHAFTS SUBJECTED TO COMBINED BENDING AND 

TWISTING. 

1991. A good example of a shaft coming under this 
head is an engine crank-shaft carrying a heavy fly-wheel. 
Let B = bending moment ; 
T = twisting moment; 

T^ = the twisting moment, which would have the 
same effect as B and T acting together. 



Then, T^ = B + i/B' ~\- T\ (255.) 

The twisting moment T^ is called the ideal twisting 
moment, and should be used in formula 2>S2> to deter- 
mine the diameter of the shaft. 

Example. — Calculate the diameter of a steel crank-shaft for an 
engine from the following data: The length of stroke is 10 feet. The 
maximum tangential force acting on the crank-pin is 25 tons. The 
shaft is 10 feet long between centers of bearings, and carries midway 
between bearings a fly-wheel weighing 55 tons. Assume the safe 
stress to be 9,000 lb. = 4^ tons. 

Solution. — The twisting moment T= maximum tangential force 

X length of crank = 25 tons X 5 feet = 125 foot-tons = 1,500 inch-tons. 

65 V 10 
The bending moment = ^ P / = — J"^ = 137.5 foot-tons = 1,650 

inch-tons. 

Then, Ti = B+ xT^TT* = 1,650 + |/l,6508-f 1,500* = 8.880 inch- 
tons. 
Now, using formula 252, 

^=1.72j/^ = 1.72.j/^ = 1.72^2^ = 16f, nearly. Ans. 

It will be noticed that, in the above solution, the units 
selected are inches and tons. Generally, when the bending 
and twisting moments are large, it is more convenient to 
use tons than pounds. Care must be taken, however, to 



MACHINE DESIGN. 



1279 



make the units consistent ; for example, it would have been 
wrong to have taken the moments in inch-tons^ and the safe 
stress in pounds. 

1992. It is sometimes convenient to put formula 255 
in another form. 

■D 

Let -Tp^ Z\ then, formula 255 becomes 



r, = z r+ |/z^T^"+^ = T{z-\- i/z^ + i). 

Let Z+ s/Z"" 4- 1 = k\ then, T^ = k T. 

But, by formula 252, the diameter is proportional to 
the cube root of the twisting moment. Therefore, if d is 
the diameter of shaft required by the twisting moment 7", 
and d^ the diameter required by the twisting moment T^ , 

-^ = y -^ = \^, or d^ = d^. 

The following table gives the values of 4/^ for varying 

values of Z: 

TABLE 49. 



z 


f /^. 


Z 


fl. 


Z 


^~k. 


.2 


1.068 


1.2 


1.403 


2.2 


1.665 


.4 


1.139 


1.4 


1.461 


2.4 


1.710 


.6 


1.209 


1.6 


1.516 


2.6 


1.753 


.8 


1.277 


1.8 


1.568 


2.8 


1.794 


1.0 


1.341 


2.0 


1.618 


3.0 


1.833 



This table may be used advantageously for computing the 
diameter of shafts subjected to twisting and bending. The 
diameter is first computed for twisting alone, and is then 
multiplied by the value of y~k corresponding to the ratio Z 
between the bending and twisting moments. 

Example. — A wrought-iron shaft transmits 150 horsepower at 125 
revolutions per minute, and is at the same time subjected to a bending 
moment of 60,000 inch-pounds. Calculate the diameter of the shaft 

Solution. — For twisting only, 



^=3.62 



y H 



N 



= 3.62, 



See Art. 1416. 



1280 MACHINE DESIGN. 



By formula 231, T=PR=z 63,025 ~ = 75,630 in. -lb. 

B = 60,000. 

-. B ^ 60,000 Q . 

Then, -^ = Z= t^^-^ = .8, nearly. 

From the above table ^^= 1.277. 

Hence, d= 1.277 X 3.62.1/^ = 444'. Ans. 

r 125 



HOLLCW SHAFTS. 

1993. It is plain that the outer fibers of a shaft are 
much more useful in resisting a twisting or bending strain 
than the fibers near the center ; hence, if a shaft is made 
hollow, its weight will be diminished in much greater pro- 
portion than its strength. In other words, a hollow shaft is 
stronger than a solid one containing the same amount of 
material per unit length. 

Let d^ = outside diameter of hollow shaft ; 

d^ = inside diameter of hollow shaft ; 

d = diameter of solid shaft having tne same strength 

as hollow shaft. 

SI 
The resisting moment (see Art. 1 398) is . Taking 

the axis of the shaft as the neutral axis, / becomes J"; the 
values of / for both hollow and solid shafts are double those 
given in the Table of Moments of Inertia, where the neutral 
axis is the diameter of a cross-section. 

S,nd* 



For a solid shaft, ^ = -^ = ^4^. 
c d 16 

s^^{d:-d:) 

For a hollow shaft, ^= ^^ = ^il!^^^L<l 

c d^ 16^, 

2 



MACHINE DESIGN. 1281 

Hence, if the shafts are to be equally strong, 

16 16 < 



or 



'=^=-.'('-» 



1 

Let -^ = in. Then, ^' = ^/ (1 - m% 

or ^^ = ^|/--^. (256.) 

Example. — Suppose we wish to find the diameter of a hollow shaft 
which will have the same strength as a solid shaft 8 inches in diameter, 
assuming the hole to be one-half the diameter. 

Solution. — Applying formula 256, in =-^, and 

di = d i/ ^ = 8 i/ 1 = 8.174, say 8^^" outside diameter, 

f 1 — in^ r 1 — {^Y 

and A:^^" inside diameter. Ans. 

Hollow steel shafts are muchr used for marine engines. 



SHAFT COUPLINGS. 

1 994. Couplings are used to connect the ends of shafts, 
and are of three kinds: 1, Fast or permanent couplings. 
2. Loose couplings, or clutches, by means of which shafts 
may be connected or disconnected at pleasure. 3. Friction 
clutches, which are loose couplings that hold by friction. 

1995. Box, or Muff, Couplings. — This coupling, as 
shown in Fig. 671, consists of a cast-iron cylinder which fits 




Fig. 671. 
over the ends of the shaft. The two ends are prevented 
from moving relatively to each other by the sunk key. The 
key-way is cut half into the box and half into the shaft 



D. 0. III.— 21 



1282 



MACHINE DESIGN. 



ends. Quite commonly the ends of the shafts are enlarged 
to allow the key-way to be cut without weakening the shaft. 
The key may be proportioned by the formulas already 
given. For the other dimensions take 

t=4. d+.5'.) (*^^-) 

Example. — Find the dimensions of a muff coupling for a shaft 3^" 
in diameter. 

Solution. — For the key we use formula 233. 



For the muff, we use formula 257. 

/=2i^+ 3" = 2ix2^+ 2'' = 8r. ? 

.5" = U". ) 



Ans. 



/= .4^+. 5"= .4x2i + .5" = H" 

1996. A clamp coupling is shown in Fig. 672. The 
faces for the joint are first planed off, the holes are drilled, 
and then the two halves are bolted together with pieces of 




Fig. 672. 

paper between them, after which the coupling is bored out 
to the exact size of the shaft. The pieces of paper upon 
being removed leave a slight space between the halves, and 
the coupling when bolted to the shaft grips it firmly. 

This form of coupling is very easily removed or put on; 
it has no projecting parts, and may be used as a driving 
pulley, if desired. The key is straight, and fits only at the 
sides. 

The following are the proportions used in practice ; 

d= diameter of shaft; 

Diameter of coupling D = ^^d-[-Y ) 

Length of coupling 1= ^ d. 

The diameter of the bolts may be f for shafts under %Y 
in diameter ; f" for '^\" shaft, and ^" for larger shafts. 



MACHINE DESIGN. 



1283 



For shafts up to 3 inches in diameter use 4 bolts; for larger 
shafts use 6 bolts. 

1997. The flange coupling is shown in Fig. 673. 
Cast-iron flanges are keyed to the ends of the shafts. To 
insure a perfect joint the flange is usually faced in the lathe 
after being keyed to the shaft. The two flanges are then 
brought face to face and bolted together. Sometimes the 
ends of the shafts are enlarged to allow for the key-way. 




Fig. 673. 

To prevent the possibility of the shafts getting out of line, 
the end of one may enter the flange of the other, as shown. 
The following proportions may be used for the form of 
flange coupling shown in Fig. 673: 

d = diameter of shaft. 
D =lf^+r; 
Diameter of bolt circle D^^'l^d^ 2"; 

Number of bolts n — Z -\- — \ 

,1 

(Take the nearest even number. ) 



Diameter of bolts d, = 



d 



n 



A-U' 
^ 4 



Diameter of coupling D^ = 1AD^\ 

b =i^+V 
e =%b\ 



(258.) 



1284 



MACHINE DESIGN. 



The key may be proportioned by the formulas already 
given for keys. 

1998. Solid flange couplings, shown in Fig. 674, 
are used largely to connect the sections of propeller shafts. 




Fig. 674. 

The flanges are forged on the ends of the shafts, and are 
connected by tapered bolts. 

Let n = number of bolts uniting flanges ; 

d= diameter of the bolts at the joint formed by the 

flanges ; 
£> = diameter of shaft; 
B = diameter of bolt circle. 



The shearing resistance of the bolts is 



TT^ 



, nS ^,.a.nd the 
4 " 



moment of this resistance about the center of shaft is 



MACHINE DESIGN. 



1285 



Now, it was shown in the derivation of formula 253 
that the moment of resistance of the shaft when subjected 

to twisting is -^rw-S^. If the bolts and shafts are of equal 



strength. 



16 



Tzd'nS.B TzD'S. 



or 



d'^nB^ 



16 * 



2* 



Approximately, j5= 1.6 Z^. Using this value, we obtain 

But more nearly, B ^^ D -\-%d. 

Substituting for ^its value just found, 

2 



^ = ^(i + 7fc)- 



Whence, ^'^ 2? (l + -^)^f. 



or dz^D 



/ A 



2;/ 



2;<4/3.2/2 + 2) 
The values oik = V ^^^ 



kD (259.) 



of n are as follows: 



2;^(|/3.2;2 4-2) 



for different values 



TABLE 50. 



n = 


3 


4 


5 


6 


7 


8 


9 


10 


k = 


.318 


.283 


.258 


.239 


.224 


.212 


.201 


.192 



1286 



MACHINE DESIGN. 



The following are the proportions of the other parts of 
the coupling : 



C =D-\.4id; 



(260.) 



The bolts are tapered f inch per foot of length. 
1999. Seller's Cone Coupling.— This coupling is 
shown in Fig. 675. It consists of an outer box or muif 




FIG. 675. 

which is cylindrical externally, but has the form of a double 
truncated cone on the inside. Within the muff are placed 
two slotted sleeves, which are turned on the outside to fit 
the muff, and also bored out to fit the shaft. These sleeves 
are pulled together by three bolts, and as they are drawn 
farther into the muff, they grip it and the shafts firmly. 

The bolt holes pass through both the sleeves and muff, 
and are square in cross-section. The friction between the 
sleeves and shaft is generally sufficient to prevent slipping, 
but to be on the safe side, the sleeves are usually keyed to 
the shaft. The key should have no taper, and fit at the 
sides only ; its proportions may be obtained by formula ;233, 

The other proportions may be taken as follows: 
d = diameter of shaft ; 
l = 4:d) d^= id;\ 
D=dd; e=lid;[ (261.) 

A=:id. ) 



MACHINE DESIGN. 



1287 



The conical sleeves may be tapered 4 inches per foot of 
length. In putting up lines of shafting, the couplings 
should be placed, if possible, near bearings, and on the side 
of the bearing farthest from the driving point. 



2000. Universal 
•Joints, or Flexible 
Couplings. — The princi- 
pal forms of these joints 
have already been described 
in Arts. 1458, 1459, and 
1460. 

In Pig. 676 is shown a 
common form of these joints 
which, when constructed of 
wrought iron, may have the 
following proportions : 

d^ diameter of shaft; 
^ = 1.8^; e—\.^d\ 
b—%d', g=.Qd; 
c = d\ h= .^ d; 

k— .^ d. 




Fig. 676. 



(262.) 



2001. Loose Couplings. — These couplings are used 



4^ 



n 



K- c-*we^ 



:"J 



Ai-A 



a d 




Fig. 677. 



1288 



MACHINE DESIGN. 



when the shafts are to be alternately connected and dis- 
connected. For large slow-moving shafts, the claw coupling 
shown in Fig. 677 may be used. This coupling somewhat 
resembles the flange coupling, Fig. 673, except that the 
flanges, instead of being bolted together, are provided 
with a set of lugs c which fit into each other. One flange 
is permanently fastened to the shaft by a sunk key m^ 
while the other is fastened to its shaft by a feather key n^ 

and may be moved back 
and forth, thus throwing 
the coupling, or clutch, 
in or out of gear. The 
movement of the clutch 
is effected by a forked 
lever fitting into the re- 
cess h. 

The lugs or claws may 
be given in the form shown in Fig. 678, in which case the 
couplings are more easily put in gear, but will drive in only 
one direction. 




Fig. 678. 




Fig. 679. 



MACHINE DESIGN. 



1289 



Cast-iron claw couplings may have the following pro- 
portions : 

diameter of shaft ; 



d 



a = l^d; 


e=%d; 


b = 2f d; 


/i = id; 


c = § d; 


s = ^d 



t = ld 



(263.) 



2002. Friction couplings, or clutclies, are used 
as loose or disengaging couplings on shafts running at high 
speeds. They are often used to couple wheels or pulleys to 
shafts. The form shown in Fig. 679 is simple in construc- 
tion, but it is open to the objection that it is hard to put in 
gear. Besides, the horizontal component of the pressure be- 
tween the conical surfaces causes an end thrust on the shaft. 

The average diameter of the conical part may be from 
4 to 8 times the shaft diameter, according to the amount 
of power to be transmitted. The angle m of the cone may 
be from 4° to 10°. The other proportions are as follows: 

d = diameter of shaft ; 
a = 2d; e = ^d; 

d = 4ctoSd; h — \d', 

^ — 1 6 "> 



k = \d. 



(264.) 



c = 2\d; 
t=.l\d; 

A better form of friction clutch is shown in Fig. 680. 
The shaft it carries a flange or cylinder A^ and the shaft m 




Fig. 68a 



has keyed to it a ring B, The ring is split and fits inside 



1290 



MACHINE DESIGN. 



the flange or cylinder A. The split ends are connected by 
a screw with right and left hand threads. The screw is 
turned by the lever (7, which is connected by the link D to 
the sleeve E. When the sleeve is pushed towards the 
clutch, the rotation of the screw throws the ends i% F of 
the ring B apart, and thereby causes the ring B to grip the 
flange A tightly. -A clutch of this form is easy to operate, 
and produces no end thrust on the shaft. 

Another form of clutch differs from the preceding in 
having the friction band B on the outside of the flange. 
The principal proportions of these clutches are about the 
same as given above for Fig. 679. 

2003. Shifting Gear for Clutches. — A clutch is 
usually put inj^r out of gear by means of a forked lever, the 




r 



■h- 



l y 




Fig. 681. 



prongs of which fit into the groove cut in the sliding part 
of the clutch. The lever is usually worked by hand, though 
for large clutches the end of the lever may be moved by a 
screw and hand wheel. The ordinary design of the forked 
end of the lever is shown in Fig. 681. To increase the wear^ 
ing surface a strap may be used, as shown in Fig. 682. The 
Strap completely fills the groove, and is often made of brass. 



MACHINE DESIGN. 1391 

The dimensions h and s are the width and depth of the 





Fig. 682. 

groove, respectively, which are to be determined b) the 
rules already given for the proportions of the clutch. 

Proportions for clutch lever : 

d = diameter of shaft ; 

h = id; s=^d;\ 

k=^%d; m = %d; >■ 

I = A^; nz=.\d, ) 



(265.) 



MACHINE DESIGN. 

(CONTINUED.) 



BEARINGS. 

2004- Solid Journal Bearings. — The simplest form 
of bearing for a journal is merely a hole in the frame 
which supports the rotating piece. Such a bearing is shown 
in Fig, 683. Motion endwise is prevented by two collars, 




Fig. 683. 



Fig. 684. 



one of which may be forged on the shaft, and the other made 
separately and held in place by a set-screw or pin. A boss 
is cast upon the frame, as shown in the figure, in order to 
give the journal the necessary length and bearing area. 

Such a bearing has no means of adjusting to take up the 
wear ; for this reason it is better to use the form of solid 
bearing shown in Fig. 684. Here the hole is bored out 
larger than the journal, and lined with a bushing made of 

For notice of copyright, see page immediately following the title page. 



1294 



MACHINE DESIGN. 



brass or other metal. The wear thus comes on the bushing, 
which can easily be replaced. 

2005a Divided Bearings. — In many cases solid bear- 
ings are undesirable, and in others it will be impossible to 
use them. The bearing is then divided, and the parts held 




Fig. 686. 



together by bolts. When this is done, the parts of the bush- 
ing are called seats or steps. The division of the bearing 
permits its adjustment for wear. 

The load on a bearing usually acts constantly in one 
direction, and the bearing should be divided so that the line 
of division is perpendicular, or nearly so, to the direction of 
the load. Fig. 685 shows various methods of dividing the 
seats. In A and B the direction of the load is vertical; 
consequently, the seats are divided horizontally. 

Naturally, the wear will come on the top and bottom of 
the seats, and the hole will become oval, with the long 



MACHINE DESIGN. 



1295 



diameter vertical. The seats are then screwed closer to- 
gether until the hole regains its original circular form. 
The necessity of making the division perpendicular to the 
direction of the load is thus apparent. In C the direction 
of the load is oblique, and the adjustment perpendicular to 
it ; in Z^ the direction of the load is oblique also, but the 
seats are divided to allow a horizontal adjustment. 

A simple form of divided bearing is shown in Fig. 686. 
It is of cast iron, with no separate seats, but with means of 




Pig. 686. 

adjustment. The bearing forms part of the frame of the 
machine. This form of bearing is used only on cheap work. 
Sometimes, however, it has a recess, or groove, cast in it, 
which is filled with babbitt metal, upon which the journal 
rests. 

The proportional unit for this bearing is 

The dimensions / and d of the bearing are the same as 
those for a journal; the other proportions are given in terms 
of the unit s^ and are shown in the figure. 



2006. Seats, or Steps. — Carefully made bearings are- 
always lined with so-called seats^or steps^ which are made 
of brass, gun-metal, phosphor-'bronze, white metal or bab- 
bitt metal, although other aiPloys are ajl,sq ig;isfcd. When. 



1296 



MACHINE DESIGN. 



made of a metal resembling brass in color, seats are often 
called brasses. The seats when worn out may be easily- 
replaced, and being made of softer material than the journal, 
the latter wears but very little. 

An ordinary form of seat is shown in Fig. 687. A part 
of the seat at each end is made octagonal in cross-section. 
This part is fitted into an octagonal recess in the pillo^v- 
block, or pedestal, which holds it, and the seats are thus 
prevented from turning. Sometimes the octagonal parts 




Fig. 687. 

are dispensed with, and the seats are turned in a lathe, the 
pillow-blocks, or pedestals, being bored to receive them. It 
is then necessary to provide the seats with a lug, or pin, to 
prevent them from turning. 

These seats may be made according to the following pro- 
portions, in which </, the diameter of the journal, is the 
proportional unit; 



b = -^d~\- 



3 " 



k=\\d. 



e = kd-\-^', 

f=^-d. d = diameter of journal. 

The dimension a should be made to fit the pillow-block 
for which the seats are intended. 

Seats should be well supplied with grooves and channels, 
so that oil may be conducted to every part of the journal. 



MACHINE DESIGN. 



1297 



It is a good plan to place a groove parallel to the axis of 
the journal in the upper half of the seat, where it may be in 
direct communication with the oil holes. Then, the advanc- 
ing side of the journal will always carry a thin film of oil 
along with it. The grooves may be from -J" to ^" wide, 
according to the size of the journal. 

2007. Lining for Seats. — Seats for large bearings 
are often lined with babbitt metal, or antifriction metal. 
It has been found by experience that a bearing will run 



o O O) 

O O O O 



O O O O 

C^ O C5 





cooler when so lined, probably because the babbitt metal, 
being softer, accommodates itself to the journal more 
readily than the more rigid gun-metal. 

Some of the common methods of lining the seats are 
shown in Fig. 688. 

At (a), the babbitt metal is shown cast into shallow helical 
grooves; at (d), into a series of round holes, and at (c), into 
shallow rectangular grooves. Consequently, the journal 
rests partly on the brass and partly on the babbitt metal. 

In cheap work, the seats are frequently made entirely of 
babbitt metal. A mandrel, the exact size of the journal, is 
placed inside the bearing, and the melted babbitt metal is 
poured around it. In better work, a smaller mandrel is 
used. After the metal hardens it is hammered in; the 
bearing is then bored out to the exact size of the journal. 



PEDESTALS. 

2008. The names pedestal, pillow-block, bearing, 
and journal-box, are used indiscriminately. They are all 
a form of bearing, and mean a support for a rotating piece- 

V. O. III.— 24 



1298 MACHINE DESIGN. 

2009. A form of journal-box frequently used for small 
shafts is shown in Fig. 689. It consists of two parts: (1) 
the box which supports the journal, and (2) the cap which is 
screwed down to the box. In this journal-box the seats are 
of babbitt, or, as it is commonly expressed, the box is bab- 
bitted. The cap is held in place by what are called cap- 
screws — an invariable method in small pedestals. 

2010. The proportioning of a pedestal is largely a 
matter of experience. Few or none of the parts are calcu- 
lated for strength. 

All the proportions of the pedestals which follow are 
based on the diameter of the journal d as the unit; the 
length of the seats is the same as that of the journal. 

For the journal-box shown in Fig. 689, the following pro- 
portions may be used for sizes of journals from ^' to 2" 
diameter, inclusive. The diameter of shaft d = the unit. 

^==2.25^. l = .OSd. 

b=l.l^d, ;;2 = .25^+.1875". 

c =^ d. n =^ .b d. 

e = .mhd. = . 625" (constant). 

/= .08 ^-f .0625^ p = 1.6d. 

^=1.75^. ^=1.333^. 

^=2.45^. r=.08^. 

i = .?>d. s= . 125" (constant). 

j = .33 d. t = .16 d. 

k=.25d+.125\ u = 1.333 d. 
v = .125d. 

2011. In Fig. 690 is shown a common form of pedestal 
which is used for somewhat larger journals than the one 
shown in Fig. 689. 

It consists of (1) a foundation plate which is bolted to the 
foundation on which the pedestal rests ; the plate is essential 
when the pedestal rests on brickwork or masonry, but may 
be dispensed with when the pedestal rests on the frame of 
the machine ; (2) the block which carries the seats and sup- 
ports the journal ; (3) the cap which is screwed down over 



MACHINE DESIGN. 



1299 



the seats. The bolt holes in both foundation plate and block 
are oblong, so that the pedestal may be readily adjusted. 

The following proportions may be used for this kind of 
pedestal, with journals from 2^^ to Q\ inclusive. An oil cup 




having a Y pipe tap shank may be used on pedestals for jour- 
nals having diameters from S'' to 4'', and %" pipe tap shank 
for larger sizes up to 6" diameter. 

Note. — The shanks of oil cups and grease cups bought in the 
market are made with a \" , ^" , f", or ^" pipe thread. The amount of 
oil or grease the cup holds when filled is usually expressed in ounces. 



1300 



MACHINE DESIGN. 




MACHINE DESIGN. 1301 

The diameter of journal </= the unit; 

a =3.25 d. 
b=l.lbd. 
c =^ d. 
e =. .bd. 
f = AZ16d. 
£- = .Odd. 
A = .S125d. 
i = .25d. 
j =.375^. 
k = 1.0625 d. 
/=.S75d. 
m = l.'75d. 
n = 1.25d. 
^ = .125" (constant). 
/ = .875" (constant). 
q = .625d, 
r = .25d. 
s = .lS75d. 
t = .65d. 
u = .'75d. 
v=1.375d. 
x = .25d, 
y = .5d. 
z = .0Q25d. 

2012. Crank-Shaft Pedestals. — The load on a 
crank-shaft bearing is due partly to the weight of the shaft 
and fly-wheel, and partly to the alternate push and pull of 
the connecting-rod. The direction of the resultant of these 
two forces is, therefore, more or less oblique, and, 
consequently, the pedestal is often divided obliquely. 

2013. A pedestal of this kind is shown in Fig. 691, 
which may be proportioned by taking the diameter d of the 
journal as the unit, and using the following proportions: 

The line ^4 ^ is at an angle of 45° to the base line. 

This pedestal, as shown, is babbitted. The babbitt is held 



130a 



MACHINE DESIGN. 



in place and prevented from turning with the journal by pro 
viding the surfaces with which the babbitt comes in contact 



H- 



/ 



-n 



rnr-"-" 



m 



^ 



-X-* 




with round projections as shown. The projections are 
about one inch in diameter for the larger sizes of pedestals. 



MACHINE DESIGN. 1303 

a^ = 1.2d-}-l". 

b —l.hd, 

c =.22 ^+.5". 

e^ =.l^+.25". 

f = .5" (constant). 

^, =.l^+.25". 

k =.dd. 

i =.15 ^+.25". 

j =.15 ^+.375". 

k =.3^+.625^ 

/ =.l^+.375''. 

m = .25d. 

n = .3d. 

p = .25 ^+.25", 

/, = .04^. 

q =.d5d. 

r =.02 ^+.3125". 

s =.l^+.25". 

/ =.75d. 

u =.27^+1.125". 

V = 1.45^. 

w = 1.75^. 

X =.25 ^+.625". 

^ =1.3^+1". 

z =.25^+1". 
To find the radius e draw a line parallel to the base line, 
and at a distance x above the base line. The point of inter- 
section O of this line with the line y^ ^ is the center of the 
arc having the radius e. The radius g^ is found by trial. 
The center of the arc must lie on the line A B. With a 
radius ^2 +«^i describe an arc from the same center. Draw 
a line parallel to the face of the bearing at a distance from 
it determined by q. The radius g is to be found by trial, 
the center being on the base line, and the arc tangent to the 
line determined by q and to the arc determined by^2+<^i' 
The rib R may be used on all pedestals above 6" diameter of 
journal, o' may be tapped with a \" pipe tap for 3" to 4" 



1304 



MACHINE DESIGN. 



journal, %" pipe tap up to G" journal, Y ^P ^^ ^" journal, and 
f ' for all sizes above 9". 

Two oil cups may be used on all bearings above 8" diam- 
eter of journal. This style of pedestal may be used for 
journals from 3" up to 15" diameter, inclusive. 




Fig. 692. 



2014. Fig. G92 shows a pedestal suitable for the crank- 
shaft of a horizontal engine with iournals from 8" to 20" in 



MACHINE DESIGN. 1305 

diameter. The block may be complete in itself, as shown in 
the figure, but more often it forms part of the engine bed. 
The seats are in three parts, and may be adjusted horizon- 
tally by means of the wedges IV. The lower seat may be 
raised by placing packing pieces under it. To obtain its 
dimensions, use the following proportions, which are based 
on the unit d= the diameter of the crank-shaft journal: 



a ^d-\-V\ 


q' =1.5d. 


b =.5^+1". 


r =.15d. 


c — ,md. 


r' =.ld. 


e =.825 ^-.25" 


r,==d 


/ = .6^. 


s =.9^. 


^- .1^4-. 5625". 


/ =.15 ^+.375". 


h =:.l^-f .25". 


/' =,9d. 


h'=..O^d. 


u =1.5d. 


i =.lld. 


■V =.25 ^+.375". 


j — .025" (constant). 


w = 1.4:5 d. 


k =.5^-f-1.25". 


W' = 1.4:1 d. 


/ =.375" (constant). 


w^ = 1. 75 d. 


i';/ = . 175 ^-f .3125". 


X =.ld. 


n =.25^-f .25". 


y =.3^-f .75". 


^' = .l^+.375". 


y =.2 ^+.5". 


= 1" (constant). 


z = .md. 


/ =.25 ^4-. 625". 


z' =2.5" (constant) 


q =l,75d. 





Taper of adjusting wedge, 1 : 10. 

Further details of the bottom seat and the cap are shown in 
Fig. 093, in which the unit is the same as in Fig. 092, and the 
proportions are as follows: 

a = 1" (constant). c = .OSd. 

b = 1.65d—.5". d=.ld. 

2015. The foundation, or bed casting, is shown in Fig. 
694, and has dimensions to suit the pedestal shown in 
Fig. 092. Its proportions are as follows, the diameter of 
the crank-shaft journal ^ being taken as the unit: 



1306 



MACHINE DESIGN. 



a 
b 
c 
e 

f 
g 
h 



2.45^+7.25". 

2.3^+5.25". 

.5^+3.5". 

3.5^+2". 

.25 ^+.5". 

.25^+1.75". 

.25^+2.25". 



n = 

= 

o' = 

^" = 

/ = 

^ = 

r = 



1.55^+2.5". 

.25(^+2". 

.25 ^+.5". 

.5^+4.5". 

.08^. 

1.5^. 

.15 ^+.375". 



zu 



'M 



r 



jHZ 



'~^ 



\r 



'-\ 



\A 






Fig. 693. 


i =.05 ^+.5". 


s =.15^/+.375'' 


j =.05^+1.125". 


/ =.9^. 


k =.05 ^+.75". 


2^ =.15 ^+.875'** 


/ =.25 ^+.75". 


z; = .2^. 


m = Ad, 


^ = 1.5<a?'. 


m' = .Qd. 


;ir =1.65^/. 



MACHINE DESIGN. 



1307 



201 6. Ball-and-Socket Bearings. — These bearings, 
now largely used in the United States, were first introduced 
by William Sellers & Co., of Philadelphia. They have very 
long steps made of cast iron bored to fit the journal. In 
some cases, however, the steps are cast with a recess in which 
a babbitt lining is poured to form a bearing surface for the 
journal. The steps have a spherical enlargement at the 
center, which fits into corresponding hollows in the block 
and cap, thus making a ball-and-socket joint, which leaves 




Fig. 694. 

the bearing free to move slightly in any direction to con- 
form to an inequality or want of alinement in the shaft. 
When rigid bearings are used, they must necessarily be short 
on account of unavoidable deflections of the shaft due to belt 
pull, thrust of gearing, etc. Pivoted bearings, on the contrary, 
by reason of their flexibility, may be made long, thereby 
giving a large wearing surface and increased durability. 

^OIT. Fig. 695 shows a pedestal with ball-and-socket 
bearing. The steps are of cast iron, and have a length of 
four times the diameter of the journal. 

Ordinarily the journal is lubricated through the oil hole 
in the center of the cap, but the upper step also has two 
cups, filled with a mixture of tallow and oil which melts at 
about 100° F. If the bearing becomes heated this mixture 
melts, and thus helps in the lubrication. 



1308 



MACHINE DESIGN. 



The dimensions of this bearing are obtained from the 
following proportions, which are based on the diameter of 
the journal ^as the unit: 






FIG. 695. 


a =2.375^. 


y^^ = .4^+.375''. 


b =1.75^. 


/ =1.33^. 


c =.875^. 


m —d-\-.Vlh". 


d^^^d. 


n = d. 


d^=l.ld. 


n^ = . 125" (constant). 


d^ = l.25d-\-.5\ 


=.25d. 


e =1.3d. 


p — .8^. 


/ = 1.65d. 


/, = .95^. 


£- = .2d+.25'. 


A = ^. 


h =.2 ^+.375". 


A = ^- 


i =.1875^. 


q =.01d. 


y = 1.27^. 


r = Ad. 


k =.3 ^4-. 25". 


r, = Ad. 



MACHINE DESIGN. 1309 



s = .0625" (constant). 


t£/^ = 2.06^. 


/ =2.5626^. 


X =.18^+2,5". 


u = lA5d, 


;r, = .1875^. 


u^ = 1.8^. 


y =.25 ^+.375". 


V =.S'75d. 


J/, to be found by trial. 


w =.2^+.375^ 


z =1.5^+.3I25^ 



TVALIv BRACKETS AND HANGERS. 

2018. Wall Brackets, or Post Hangers.— A shaft 
must sometimes be supported by a bearing fixed to a wall 
or pillar. In such a case the bearing is generally supported 
by a ^wall bracket. It will readily be seen that the bracket 
is a cantilever with practically a uniform load, due to its own 
weight, and a concentrated load, due to the weight of the 
shaft ; hence, the bending moment diagram resembles that 
shown in Fig. 332, Art. 1391. 

Now, in order that a cantilever should be equally strong 

SI 

at all sections, the resisting moment, — , of that section 

should be proportional to the bending moment at that sec- 
tion. Suppose the bracket in question to be a plate of a 
constant thickness b. Then, at any section, the bending 

moment J/ = — = \%^ = ^Sbh\ 

Since \ S b remains constant for all sections, the Ji^ is 
proportional to the bending moment; or, in other words, 
the height of a cantilever at any section should be propor- 
tional to the square root of the bending moment at that 
section. It can be shown mathematically that, supposing 
the cantilever to have a constant width, it should have a 
parabolic form when the load is concentrated at the end, 
and it should have a triangular form when the load is uni- 
formly distributed. In practice, a bracket or other machine 
part of the cantilever class is given a form which is often 
neither exactly parabolic nor triangular, but approximates 
closely to one or the other, according to the taste and judg- 
ment of the designer. 



1310 



MACHINE DESIGN. 



2019. A good design for a wall bracket is shown in 
Fig. 696. The bearing proper resembles that of the 
pedestal, Fig. 695. The spherical enlargement at the center 




Fig. 696. 

of the steps is held between the ends of two hollow cast-iron 
plugs, which are threaded at their outer ends, and screw into 
bosses cast on the bracket. 

The bearing may be raised or lowered by screwing the 
plugs up or down. In order to turn the plugs the hole at 
the end is made square to receive a key. When the plugs 
are in the desired position they are locked by the set- 
screws. 



MACHINE DESIGN. 1311 

The length of the bearing is 4 times the shaft diamecer; 
the other dimensions are to be obtained from the figure and 
by the following proportions: 

Diameter of shaft ^= unit; 

a =Zd. k =1.875^. 

b =7.5^. / =1.125^. 

c =5.375^. m = .l^bd, 

e —Zd. n = .^6d. ^ 

/=:1.75^. o r=.1875^. ' 

g =1.0 d. p =.d76d. 

h = 2.126 d. q = .625 d. 

i =.16^. r = 1.25 d. 

j =.25d. s =.25^+.5^ [ 

Proportions of oil or grease cup covers : 

z/ = 1.125 ^+.1875\ J/ =.0625^. 

w = 1.5 d. z =d- .0625", 

X = .125" (constant). 

The proportions of boxes, bosses, plugs, drip pans, etc., 
are to be made the same as those of the shaft hanger shown 
in Fig. 697. 

2020. Hangers. — A hanger, is used when a shaft bear- 
ing is to be suspended from the ceiling. Fig. 697 shows a 
form of a hanger made by a leading manufacturing company. 

The bearing is the same as that of the wall bracket, Fig. . 
696. The frame of the hanger is divided, and the parts con- 
nected by bolts. With such a form the shaft may be more 
easily removed than when the hanger frame is a solid piece. 

The units for determining the leading dimensions of a 
shaft hanger are the diameter d of the shaft and the drop 
D of the hanger. 

The following proportions are suitable for shafts ranging 
from 1^" up to 4 J" in diameter: 

A =(j d-\-.4:5 B. £ =2d-{~ .25 D. 

A^ = 2d-\- .03 D. F =.5 d-\- .01 D. 

B =4:d-{- .35 D. F^ = 1.5 d-\- .05 D. 

C =2d-\-,Z D, G =1.25 d. 



1312 MACHINE DESIGN. 



H =%d. 


J =.25 ^+.25^ 


I =Ad 


j\ =.125 ^+.0625'. 


J =.125 ^+.01 i?. 


k = 2.2 d. 


K =.5d-\-.5\ 


/ =4:d. 


L =.25 ^+.5". 


m =1.4^+.375^ 


M = .75 ^+.6875". 


n = d. 


N =.25 ^+.375". 


=.25d. 


= 1.25 d. 


0^ =.0625^. 


(9, = .094 ^+.002 Z>. 


J> = d. 


P =.375 ^+.008 Z>. 


A = .0625^. 


Q =.375 ^+.008/?. 


g =Ad. 


i? and i?j (see note). 


q, =.15 d. 


5 =.25^+.005Z>. 


r =2.125 d. 


5, =.125^+.003i>. 


s =1.5d. 


T =.125^+.01Z>. 


s^=.126d. 


7"^ = (see note). 


I =2d. 


U =^d. 


/, =.5d. 


V=.5d 


t^=d. 


W = .75d. 


t^ = .^5 d. 


X =.375^. 


u = .95 d. 


Y =.25^+.125^ 


u^ = .S5 d. 


Z =.625^. 


V =.25 ^+.125'. 


^ =.15^+.375^ 


v^ = .5 d. 


^, =2.4 ^+.3125". 


w = d. 


b =.OSd. 


w^ = .125" (constant). 


c =.125^+.0625''. 


X =.25 d. 


^ =.2^. 


x^ = d. 


^, =.4^. 


x^ = A:d-\- 2'\ 


^, = .2 ^. 


y =1.25 d. 


/ =.375 ^+r. 


j^ = .75^+.0625^ 


/, =.09 ^+.25". 


j/^ = .4 ^+.0625". 


^=.75^. 


z =.06 ^+.25". 


^-, = 1.3125 ^+.125". 


^, = .12 ^+.75". 


h =1.25 ^+.1875". 


z^ = .3125" (constant). 


i =.ld. 




Thread of plugs, .5" pitch 


for all sizes. 


Note. — To find i?i, draw the 


2i.vcJ\ also, draw the arc Q tangent 


to P; then, draw a straight line 


tangent to these arcs, and i?i will be 







N=diam. of bolt 




Section 1, 2. 



L — o — J 

Section 5, 4. 




MACHINE DESIGN. 



1313 



the distance along the center line determined by B included between 
this tangent and the upper face of the hanger. Having found i?i, 
make R equal to it. 

The radius Tx is made equal to f the thickness at the middle. 

The steps of the ball-and-socket bearings shown in Figs. 
G95, 696, and 697 are of cast iron, and are bored to fit the 
journal without lining or brasses. The ball, and the 
recesses in the ends of the plugs, into which the ball is 
fitted, should be faced. The screw threads on the plugs. 
Figs. 696 and 697, may be cast on the plugs or turned, the 
latter being preferable. It is customary to use two threads 
per inch for all sizes of plugs. 

2021. Pivot or Foot-step Bearings. — These bear- 
ings are used to support the ends of vertical shafts. An 
ordinary pivot bearing is shown in Fig. 698. The end of 
the pivot rotates 
on a disk A, which 
may be of steel, 
brass, or bronze. 
The brass bush B 
prevents the pivot 
from moving side- 
wise. The end of 
the pivot should 
be of steel, and it 
may be flat on the 
end or slightly cup fig. 698. 

shaped. The proportions are given in terms of the diameter 
d of the pivot as a unit. 

Its proportions are as follows : 

ci = 2d. j=Ad. 

b = 1.6d. k=.^d. 

c=z.25d+.375\ I =.15d. 

e = 1.25d. m = 1.6d. 

/=.2^+.125'. n =.2 ^+.25'. 

^=1.75^. o =,l5d. 
^=lAd. 




D. 0. III.— 23 



1314 



MACHINE DESIGN. 



2022. Pivots witli Loose Disks. — ^When pivots are 
required to run at great speed, the relative speed of the sur- 
faces in contact may be reduced by- 
placing a number of loose disks between 
the pivot and the foot-step, as shown in 
Fig. 699. 

If <^ = the number of disks and n = 
the revolutions per minute of the pivot, 
then the relative speed of any of the 




7t 



revolutions 



Fig. 699. 



surfaces in contact is 

Suppose, for example, a 



per mmute. 

pivot runs at 3,000 revolutions per 
minute, and it is desired to reduce 
the speed between adjacent surfaces to 750 revolutions 

per minute; then, 750 = -^——, or a = 3, the number of 
^ ' a -\-r ' 

disks required. The pivot and disks may be lubricated 

by a channel, as shown in the figure. Tlie disks may be 

made of gun-metal or phosphor-bronze. Proportions are 

given in the figure. 

TOOTH GEARING. 



SPUR AND BEVEL GEARS. 

Note. — Before reading this section, the student should carefully 
review the portion of Applied Mechanics which relates to gear teeth. 

2023. Materials of Gearing. — Gearing is ordinarily 
made of cast iron. If great strength is required steel may- 
be used. Gears which are called upon to resist shocks may- 
be made of gun-metal or phosphor-bronze. Fast-running 
gears are sometimes made with wooden cogs^ or of rawhide^ 
or fiber ^ instead of metal. 

Gear-wheels are either cast with teeth and all complete, 
or with a blank rim in which the tooth spaces are afterwards 
cut with a milling cutter. Gears with cast teeth are less 
expensive; those with cut teeth are more accurate. Gear- 
cutting machines are now made, however, which can easily 



MACHINE DESIGN. 



1315 



cut teeth on both spur and bevel gears of the largest size ; 
hence, cut gears are quite generally used, except on very 
large or rough work. 

STRENGTH OF GEAR TEETH. 

2024. A gear tooth may be considered as a cantilever 
whose length is /, depth, Y, and breadth, 
b, see Fig. 700. 

Let / = the pressure acting at the 
pitch line, and, as an extreme case, let 
the total pressure come on the edge 
of one tooth, as shown in the figure. 
(Usually the pressure will be divided 
among two or three teeth.) 

Then, the bending moment =/ /. fig. 700. 




The resisting moment = — = S^ 



Sbf 



\t 



Therefore, 






{a) 



The thickness t at the root of the tooth may be taken 
equal to ^ (7, and l^^.^ C from the proportions given in 
Table 34, Art. 1555. C •= the circular pitch. 

Substituting these values in equation {a\ we obtain 






Or, 



■bC=lQ.^^, 



(266.) 



The value of the working stress 5 depends upon the 
velocity of the pitch circle of the gear, since, as the speed 
increases, the teeth become more liable to shocks. 

Let V = velocity of a point in the pitch circle in feet per 
minute. Then, if v is equal to or less than 100, take 5=4, 200 
lb. per sq. in. for cast iron. 

If V is greater than 100 feet per minute, the safe stress S 
may be obtained from the following formula : 

9,600,000 



5 = 



z^-t-^,160' 



(267.) 



1316 MACHINE DESIGN. 

To find the pitch C from formula 266, the relation be- 
tween b and C must be known. The following proportions 
are those generally used in practice: 

For wheels moving slowly or intermittently, as in hoisting 
apparatus 

^ = 2 (T to 2i (f. 
For more rapidly moving cast gears, for example, 
transmitting gears, 

b-^C \.o ?>C. 
For gears moving more rapidly, and with cut teeth, 

b=.Z C\.o ?>\C. 
For very rapidly moving gears with small pitch, 

^ = 3|- (f to 4 6^. 

Example. — A cast gear 3 feet in diameter makes 40 revolutions per 
minute, and transmits 16 horsepower. Determine the circular pitch of 
the teeth, and number of teeth. 

Solution. — R — radius of gear = 18 inches. 

From formula 231, Art. 1963, 

^ 63,025 H 63,025 X 16 . ,^^ ,^ 

P = "F, = TK T^ = 1,400 lb. 

^ nR 40 X 18 

From formula 267, 

^ 9,600,000 9,600,000 o r^o. n. 

^=ir^^M = 3^x40 + 2.160 = ^'^^^ ^^' P"" ^^- ^^- 

Hence, from formula 266, 

AT ift«^ 16.8 X 1,400 «« 

Assume d = 2iC. Then, <^ C = 2i C^ = 6.2 sq. in., or C= 1.575 in. 
Taking C= 1.5708 in., the diametral pitch is exactly 2, and the 
number of teeth 36 X 2 = 72. Ans. 
Hence, C= 1.5708. Ans, 

Number of teeth iV= Z> P = 36 X 2 = 72. Ans. 
Breadth of tooth d = 2iC= 3.93", say 3|f ". 

In calculating cast-steel gears, or gears with wooden cogs, 
the above method is applicable, but the value of 5 should 
be multiplied by 3 J- for steel teeth, and by .6 for wooden 
cogs. 



MACHINE DESIGN. 



1317 



2025. \Vear of Teeth. — Though gear teeth may be 
amply strong, they may, if poorly designed, wear rapidly. 
To ensure durability, the dimensions of the tooth should 
fulfil the following conditions: 

Let / = pressure acting at pitch line ; 
b = breadth of tooth ; 
n = number of revolutions per minute ; 
H = horsepower transmitted ; 
D = diameter of gear in inches. 



Then, 



^ = 28,000, or d = 4^^. (268.) 



In the example above, we have 4|- yy = 



H UX 16 



36 



= 2 in., 



and since the breadth calculated for strength is 3^", the 
gear will wear well. It is usually necessary to apply formula 
208 to pinions of small diameter. 



2026. Shrouded Gear Teeth.— The teeth of a gear- 
wheel are said to be shrouded when the rim is made wider 






Fig. 701. 

than the tooth, and carried outwards so as to unite the ends 
of the teeth. 

Three methods of shrotrding are shown in Fig. 701. At 
a the shrouding is carried on both sides to the end of the 
tooth. In this case it is evident that only one of the pair of 
wheels gearing together can be shrouded. At d the shroud- 
ing is carried to the pitch line, and, consequently, both 
wheels of the pair may be shrouded. At c the shrouding is 
carried up on one side only. 

Pinions with few teeth are most benefited by shrouding, 
since the teeth in that case are weak at the roots, and also 
because more wear comes on the pinion. 



1318 



MACHINE DESIGN. 



PROPORTIONS OF SPUR AND BEVEL GEAR 'WHEELS. 

2027. Rims. — Various rim sections are shown in 
Fig. 702. It is good practice to make the rim thickness c 
equal to the tooth thickness on the pitch circle for gears 




Fig. 702. 



whose circumferential pitch is greater than \\ inches. For 
gears with small pitch, we may take 



^=.4c-+r=^+r. 



(269.) 



Example. — Calculate the thickness of the rim of a gear-wheel 
whose diametral pitch is 4. 

Solution. — By formula 269, 

5 






4.P 



+ i' = A + i = i^''. Ans. 



2028. The rim of a bevel wheel may have the propor- 
tions shown in Fig. 703, where c is to be computed by the 

rules given in the last article. 

2029. When wooden cogs are 
mortised into a gear-wheel, the rim 
may have the proportions shown in 
Fig. 704. 

Usually only one of the wheels in 
a pair that gear together is fur- 
PiG. 703. nished with wooden teeth, or cogs, 

in which case the wooden cogs may be made \\ times as 
thick as the iron, teeth meshing with them; that is, 

for the wooden cogs, / = .6 C\ 

for the iron teeth, / = .4 (7, or less. 




MACHINE DESIGN. 



1319 



Fig. 704 shows two common methods of fastening the 
cogs to the rim. 

b M 



.48P 





Fig. 704. 



2030. Arms of Wheels. — The form of the cross- 
section of the arms of a gear-wheel depends upon the form 




Fig. 705. 

of rim used. The forms in ordinary use are shown in Fig. 
705. The cross-shaped form A is the one mostly used for 
spur gears; and the section shown at B, for bevel gears. 
The section shown at C is good for heavy gears, while the 
oval form Z^ is a neat form for light gears. 

In all of the above forms the dimension a lies in the plane 
of the wheel, and the dimension d at right angles to this 
plane, or parallel to the axis of the wheel. 

In calculating the strength of the arm, the ribs A^ A are 
not taken into account, as they are intended to give the arm 
lateral stiffness, and not to add strength. 

Let ^ = number of arms in wheel ; 

a = width of arm, measured at center of wheel; 

X = thickness of arm ; 

S^ = allowable safe stress in arm; 

R = radius of wheel in inches ; 

/ = pressure between gears at pitch line. 



1320 MACHINE DESIGN. 

We may consider each arm as fixed at the center of the 

wheel, and free at the end, which, though not strictly true, 

is on the safe side. Then, the total bending moment is/>R 

p R 
inch-pounds, and the bending moment on each arm is — — . 

z 

The moment of resistance is 



9 / -^-a^ X 



^a 



Hence, — — ^^\S^a^ x. 

z 

To avoid strains and breakage in casting, all parts of the 

wheel should be of nearly the same thickness; therefore, 

the thickness of the arm may be made about equal to the 

thickness of the tooth or of the rim. 

That is, X ■=^\C. 

But, C = i||/. (See formula 266.) 

Hence, ^r = |- (7 = , ^ ; 



or, 



pR _ \ASja^ 
z ~ bS ' 



pRbS RbS 
whence, a" = / . .. -- = ^ . ^ . 

Owing to the initial strain set up in the contraction of the 
casting, S^ should have a low value, say 3,000 lb. S may 
have the value given above; that is, 4,200 lb. 

i^^X 4,200 Rb 



Then, ^' 



1.4 X 3,000 X ^ ^ 



Rb 
or. a 



= V- 



(270.) 



For a case in which the proportions given by formula 270 
do not give an arm whose appearance is satisfactory, the 
value of X may be varied a little in either direction, and a 
new value of a may then be calculated so as to give the 
required moment of resistance. 



MACHINE DESIGN. 1321 

Example. — A gear-wheel 2| feet in diameter, with six arms, has a 
pitch of 2 inches, and the breadth of tooth is 5 inches. Find the 
dimensions of the arm. 

Solution. — Use formula 270. 

'' = i/^ = i/^^^f^=V^=^V, nearly, Ans. 
x = iC=l". Ans. 
For arms of oval or elliptical cross-section, the moment 

Vl . ^ 64 
of resistance is c > which equals j . (See Table 



a 



of Moments of Inertia). 

Hence, M = A^. 

' <^ 64 

Combining this equation with formula 266, we obtain, 
after reducing, 



a = 1.2 
Or, ^ = 1.75 



^/FC^ 



z 



Pz 



(271.) 



Example. — A gear-wheel 12 inches in diameter has 48 teeth and 4 
arms. The rim is 2^ inches wide. Supposing the arms to be of ellip- 
tical cross-section, find their dimensions, the thickness being half the 
breadth. 

Solution. — Using formula 271, 



a 



= l,75fg = 1.75-j/||| = 1.70-. 



a z= Iff". Ans. 



•^ ^ = -I", nearly. Ans. 

Formulas 270 and 271 give the width of arm measured 
at the center of wheel. The arm is tapered from center to 
circumference. For small gears the taper may be -3^2 on 
each side ; for larger gears, -j--^ on each side. The thickness 
X remains constant, but in the elliptical form the arm is 
tapered in both width and thickness, so that the latter is 
constantly equal to half the former. 

The average thickness of the stiffening ribs A^ Fig. 705, 
may be .4 C. The ribs are tapered slightly to allow the 
pattern to be easily drawn from the mold. At the hub the 



1322 MACHINE DESIGN. 

width of the rib may be b^ the same as the length of the 
tooth. The rib is tapered so that at the rim the width is 
lb\.o\ b. 

The number of arms to be used in a given gear-wheel is 
largely a matter of judgment. Reuleaux gives the following 
formula : 

z=.h^\fJPC, (272.) 

where z is the number of arms ; N^ the number of teeth, and 
C^ the circular pitch. 

Example. — How many arms should be given a gear 4 feet in diam- 
eter, diametral pitch 1^ ? 

Solution.— A^= 4 x 13 X li = 'J'3. 

Hence, from formula 372, 

^ = .55^/722x2.09 = 5.61. 
Therefore, use 6 arms. Ans. 

If the formula gives an odd number of arms, the nearest 
even number may be used, if desired. 

2031. Hubs or Naves of Gear-TVlieels.— The 

thickness of the hub is often made equal to the radius of the 
shaft on which the wheel is placed. If the shaft is enlarged 
for the wheel seat, the thickness of the hub is made equal 
to the radius of the main portion of the shaft. Then, if d 
represents the diameter of the shaft, 1.2 //= the diameter of 
the enlarged portion, and \d v^ the thickness of the hub. 

The above rule is very generally used, and gives good 
proportions when the wheel transmits the full power of the 
shaft. If, however, a gear transmits only a fraction of the 
power of the shaft, the hub thickness may be calculated 
from the following formula, which may also be used for 
finding the thickness of the hub of any gear-wheel: 



w = \\/bCR, (273.) 

where w is the thickness of the hub, and b^ C, and R the 
same as in the previous formulas. 

The length of the hub varies from b to lA b. 



MACHINE DESIGN. 



1323 



The hubs of large and heavy wheels may be split as 
shown in Fig. 706. This relieves the hub from the initial 
strains due to unequal contraction in cooling. 

Strips of metal are placed in the slots, and the segments 




Fig. 706. 



are held together by iron or steel bands shrunk on. The 
proportions are as follows: 



/, = f z£/; /, = f w; d = -j\w. 



2032. Built-up Wheels. — For convenience in cast- 
ing, and also for convenience in transportation, large and 
heavy wheels are made in sections, which are assembled and 
bolted together. The division of the wheel may be made 
in various ways. The hub and arms may be cast separately, 
and the rim in segments; the hub and arms may be cast 
together, and the rim in segments ; or, finally, each division 
may include a portion of the hub, an arm, and a segment of 
the rim. Fig. 707 shows one method of bolting the rim, 
segments, and arm. 

Fig. 708 shows a method in which the rim and hub of the 
gear are parted and held together by bolts. 



1324 



MACHINE DESIGN. 



Example. — Compute the leading dimensions for a spur wheel to 
work under the following conditions: 
Diameter of pitch circle — 5 ft. 6 in. 
Revolutions per minute = 45. 
Horsepower transmitted = 240. 




inn RED 




Fig. 707. 




Fig. 708. 



Solution. — The velocity of a point in the pitch circle is irDn 
= 3.1416 X 5i X 45 = 778 ft. per minute. 
Hence, from formula 267, 
9,600,000 



S= 



— = 3,270 lb. per sq. in., in round numbers. 



778 + 2,160 
From formula 231, Art. 1963, 

^ _ 63,025 A^ _ 63,025 x 240 
~ 33 X 45 



= 10,186 lb. 



Now, from formula 266, 

^^ taoP 16.8x10,186 ^^_ 
dC= 16.8 ^ = ' = 52.3 sq. m. 

Assume If = SC; 
Then, 3 C2 = 52.3; 
C = 4.17". 
circumference of wheel 3. 1416 X 66 



Number of teeth = 



4.17 



4.17 



= 49.7, 



Taking 50 for the number of teeth, we have 
3.1416 X 66 



C = 



50 



= 4. 147 inches. 



MACHINE DESIGN. 1325 

Breadth of face = ^ = 3 C = 3 x 4. 147 = 13.44 = 12yV". Ans. 
Thickness of tooth = / = .475 C= .475 X 4.147 = 1.97" Ans. 
Height above pitch line = .3 C= .8 X 4.147 r= 1.24". Ans. 
Depth below pitch line = .4 C = .4 X 4. 147 = 1.66 '. Ans. 
For the size of the shaft we may use formula 123, Art. 1415. 

^=4.92 \f^= 4.92 i/5^ = 7i" , nearly. Ans. 
J n r 45 

The enlargement for the wheel seat will then have a diameter of 

1.3^=1.2x7.5 = 9". Ans. 
For the thickness of the hub we use formula 373. 



w = i jfFC~R = i -^12.44X4.147X33 = 4", nearly. Ans. 
The length of hub may be 

1.4 ^ = 1.4 X 13.44 = 17yV', nearly. Ans. 
For the number of arms, formula 373 gives 

z = .55 i^jv^ = .55 ^ 50'-^ X 4.147 = 5.5. 
Therefore, use 6 arms. Ans. 
For the width of arm at the center of the wheel, formula 370 gives 

a = i/M = a/EEEE = 8.27, say 8i". Ans. 
f 2 r 6 

Supposing the arms to taper -^ on each side, the width of arm at 
pitch line is 

8i-(3x33xA) = 6r. Ans. 

The thickness of the arm is i C = i X 4. 147 = 3. 07 = 3yV"- Ans. 
The thickness of the stiffening rib is .4 (r= .4 X 4.147 = 1.66 = 1^". 

Ans. 

2033. Proportions of Bevel Gears. — The rules and 
formulas used in designing spur gears are equally applicable 
to bevel gears. For the circular pitch (7, used in spur-gear 
formulas, should be substituted the mean circular pitch of 
the bevel gear; that is, the circular pitch measured at the 
middle of the tooth face. 

Denote this mean circular pitch by the symbol Cy^. Then, 
the bevel wheel may be designed by formulas 266 to 273 
by placing C^ for (7, and P^ for P. 

Then, ^(f^^lG.S^ (266^.) 



5* 

4P 



.4C« + F = i4+r. (269^.) 



1326 



MACHINE DESIGN. 
{270a.) 



a = y — . 






a 



= 1.751/ • 



bR 



^=.55fF^. 



W::^.\^/bC^R. 



{271a.) 

{272a.) 
{273a.) 



^WORM GEARIIVG. 
2034. Construction of Worm Gearing. — Worm- 




Fig. 709. 



wheels may be constructed in various ways. The wheel may 
be similar to a spur wheel, except that the teeth make an 



MACHINE DESIGN. 1327 

angle with the axis of the wheel equal to that of the threads 
of the worm measured at the pitch diameter. (See Fig. 709.) 
For careful and accurate work, the wheel is cut by a 
special milling tool called a hob. This hob is a steel worm 
which has been cut in a lathe, notched to form a milling 
cutter, and then hardened and tempered. The teeth cut by 
this hob must evidently have the correct form to gear with 
a worm having the same pitch as the hob. When the wheel 
teeth are constructed by this method, there is much closer 
contact between the worm and the wheel. Therefore, the dur- 
ability and efficiency of the mechanism are largely increased. 

2035. Friction of 'Worin Gearing. — Since the 
threads of the worm slide on the teeth of the wheel, there is 
considerable work expended in overcoming friction. It, 
therefore, becomes necessary to consider the form of worm 
and wheel which will work with least friction. 
Let Rj^ = radius of pitch circle of wheel ; 
R =■ radius of pitch circle of worm ; 
Jf = the axial pitch of the worm ; 
/j = circular pitch of wheel, and divided axial pitch of 

worm; 
IV = number of threads in worm ; 
iVj = number of teeth in wheel ; 

P = resistance to turning at circumference of worm- 
wheel ; 
Q ~ actual force acting at pitch line of worm neces- 
sary to turn worm; 
Q' = force required to turn worm if there were no 

friction between it and wheel; 
e = efficiency of worm gearing; 

a = angle between tangent to worm thread and any 

plane perpendicular to axis of worm. 

The efficiency of the gear is the ratio of the useful work to 

the total work. Assuming the coefficient of friction to be about 

.15 or .16, the efficiency may be obtained by the following 

formula : 



1328 MACHINE DESIGN. 

Hence, for maximum efficiency, the radius R of the pitch 
circle of the worm should be made as small as possible, and 
the number of threads should be as large as possible. 

Example. — If the radius of pitch circle of worm is 3 times the 
divided axial pitch /i, and if the worm has 2 threads, what is its 
efficiency ? 

Solution. — From formula 274, 

^ = AT J. n = o . o J. = -40 = 40 per cent. Ans. 

If the radius R were only 2/^, the efficiency would be 

On the other hand, if the number of threads were 1, the 
efficiency would only be 



A+3A 



= 25 per cent. 



The leading problem, therefore, in designing a worm gear 
is to make the ratio ■ — as small as possible, and at the same 
time allow the worm shaft a sufficient diameter for strength. 

The twisting moment acting on the worm shaft is 

e 
From the principle of work we have 2 7tRQ' = Pp. 

Hence, 21^ = ^, 

which is the twisting moment to be used in calculating the 
worm shaft. 

TT 

Example. — The circular pitch of a worm-wheel is .7854 = -j; it has 

40 teeth, and transmits 2 horsepower at 30 revolutions per minute. 
The worm has 1 thread, and the radius of its pitch circle is 2 inches. 
Calculate the twisting moment on the shaft, and the size of the shaft, 
if made of wrought iron. 

TT 

Solution. — Diametral pitch of wheel = — = 4. 

40 1 

Diameter of wheel = -;j- = 10'. * 

4 



MACHINE DESIGN. 



1329 



Radius of wheel = i?i = 5". 
From formula 231, Art. 1963, 



P = 



63,025^ 63,025x3 



i?i n 
From formula 274, 



e — 



5x30 

.7854 



8401b., nearly. 



Np^ ^R~ .7854 + 2 



= 28 per cent. 



PP 



840 X 



= 375 in. -lb. Ans. 



Hence, twistinar moment = 7~— = -^ -kk 

* 2 7r^ 27rX.28 

Using formula 123, Art. 1415, 

diameter of shaft d—.%\ ^f/BTS = 1.364, say If. Ans. 

The pitch of the worm-wheel may be calculated from the 

formula for the pitch of spur gears. 

p 

2036. The rim and arms of the worm-wheel may be 
given the same proportions as the rim and arms of a spur 
wheel of the same diameter and pitch, except that the width 
of the face of the wheel is generally made equal to double 
the pitch. 

The rim may have the forms shown in Fig. 710. The 




Fig. 710. 
form a is best, as there is line contact between the thread 
and tooth. 

The ratio — may vary from 1 to 4 or 6. When the worm 

is cut on the shaft, the lower values may be used. When, 
however, the worm is cast and keyed on the shaft, the 



D. 0. 111. 



>k 



1330 



MACHINE DESIGN. 



ratio must be made greater. As has been shown, the 
smaller this ratio the greater the efficiency. The number 
of threads in the worm varies from 1 to 4, depending upon 
the velocity ratio to be transmitted. 

Letting N and N' represent the number of threads on 
worm and teeth on wheel, respectively, and w the velocity 
ratio, we have ^/ 



w = 



N' 



Usually it is undesirable to give the worm-wheel less than 
30 teeth ; hence, if the desired velocity ratio is less than 30, 
the worm must have more than one thread. For example, 
if the worm shaft is to make 20 revolutions while the wheel 
shaft makes one revolution, we may have either N' = 40 and 
N= 2, or N' = 60 and N= 3. 

The length of the worm is from 3 / to 6/, usually 4/; 
that is, there are usually 4 turns, as shown in Fig, 709. 



FRICTIOIV GEARING. 
2037. Friction gearing may be used for the transmis- 
sion of small powers ; it should not be used for heavy work. 
In this style of gearing the smooth faces of a pair of wheels 
are pressed together, and one drives the other by means of the 

friction between the surfaces in 
contact. 

If the resistance to turning of- 
fered by the driven wheel is less 
than the frictional resistance be- 
tween the surfaces of the two 
wheels, motion will be trans- 
mitted; otherwise the surfaces will 
slide over each other, and the gear- 
ing will not work. 

It is the usual practice to face 

the rim of one of the wheels with 

leather, wood, or paper. Fig. 711 

shows a friction wheel thus faced 

PIG. 711. with wood. 




MACHINE DESIGN. 



1331 




As shown, the grain of the wood lies along the working 
surface. When leather or paper p 

is used, the edges of the layers 
are used as the driving surface. 
The pulley which is covered with 
wood or paper must be the 
driver ; the driven pulley is usually 
of cast iron. Pi^ ^12. 

Let P (see Fig. 712) = driving force at circumference of 

friction wheels; 
Q = force acting on the bearings to 

press the wheels together; 
f=z coefficient of friction between sur- 
faces. 

Qf= P,orQ = ^. (275.) 



Then, 



/ 



The coefficient of friction f has about the following 
values : 

Metal on metal /= .15 to .25. 

Wood on metal f— .25 to .40. 

Paper on metal /"= .20. 

Leather on metal f= .25 to .30. 

The power transmitted by a pair of friction gears can be 
easily calculated. The following example shows the method : 

Example. — Suppose a friction wheel faced with leather to drive 
another 2^ feet in diameter at 300 revolutions per minute. The force 
pressing the wheels together is 450 pounds. Assuming/" = ,3, calcu- 
late the horsepower transmitted. 

Solution.— By formula 275, P = ^/= 450 x .3 = 135 lb. 



R — radius of wheel in inches = 



From formula 231, Art. 1963, 



2iXl2 

3 



= 15" 



H = 



PRN 135 X 15 X 300 



= 9.64 horsepower. Ans. 



63,025 ~ 63,025 

The calculations for bevel friction gears are more com- 
plicated, but may be easily made. 

The proportions of the rim, hub, and arm of a friction 
gear-wheel may be made the same as those of a toothed gear- 
wheel transmitting the same power. 



1332 MACHINE DESIGN. 

FLEXIBLE GEARING. 



BELT GEARING. 



BELT MATERIALS. 

2038. Leather. — The material mostly used for belts 
is leather tanned from ox hides. The leather is about -^ 
inch thick, and is obtained in strips up to 5 feet in length. 
Belts are made of any required lengthby joining these strips 
together. 

Single belts are made of one thickness of leather; 
double belts from two thicknesses of leather. 

Cotton may be used for belts which are exposed to damp- 
ness. Cotton belts can be made very wide, and without the 
many joints necessary in leather belts. The necessary 
thickness is obtained by sewing together from 4 to 10 plies 
of cotton duck. Cotton belting is cheaper and stronger 
than leather belting, but probably less durable. 

Rubber belts are made by cementing together plies of 
cotton duck with india rubber. Rubber belts are more 
adhesive than leather belts, and, hence, have greater dri- 
ving capacity. Rubber belts are considered to be the best to 
use in damp situations. 

Dimensions of Belts. — Practical formulas for compu- 
ting the dimensions of belts are given in Arts. 1483 to 
1489 inclusive. 

BELT PULLEYS. 

2039. Fig. 713 shows a solid, and Fig. 714 a split, 
cast-iron belt pulley. The general form of these pulleys 
corresponds very closely to the best modern American 
practice. The split pulley has the advantage of being more 
easily put on the shaft, especially when the shaft is in posi- 
tion or has other pulleys already on it. 

When the amount of power to be transmitted by a pulley 
is small, it may be fastened to the shaft by set screws. Split 
pulleys are also made so that the bolts through the hub will 
serve as a clamp to draw the hub tight enough on the shaft 
to prevent slipping with small loads. When the amount of 



MACHINE DESIGN. 1333 

power to be transmitted is considerable, pulleys should be 
fastened with keys, and in some cases both keys and set- 
screws are provided. 



Fig. 713. Pig. 714. 

In the rules and formulas for dimensions of pulleys, the 
following symbols will be used to represent the various 
dimensions : 

b = breadth of belt running on pulley ; 

B = breadth of pulley rim ; 

d = diameter of shaft on which pulley is keyed; 

/ = thickness of pulley rim at edge; 

a = width of arm at center of pulley; 

ze/ = thickness of hub of pulley ; 

/ = length of hub of pulley ; 

n = number of arms; 

D — diameter of pulley; 

R = radius of pulley; 

s = swell at center of pulley rim. 

All dimensions to be in inches. 

Note. — To obtain a, the arms are supposed to extend through the 
hub to the center of the pulley. (See Fig. 720.) 

2040. Rim of Pulley. — The rim is usually of the 
form shown in Fig. 715. If the rim is crowning, the curve 
may be struck with a radius of from 3 B to 5 B, in which 



1334 



MACHINE DESIGN. 



case s = — — to — -— , about. The width B is made from -1-^ 
20 40 ^ 

to f <^ ; i. e. , ^ = I ^ to f ^. 

For the thickness /, the following formula gives results 
which agree well with the practice of good shops: 



200 ^^« • 



(276.) 



For double belts, ^'^ may be added to the thickness obtained 
by the above formula. 

2041. Flange Pulleys. — When there is a liability of 
the belt frequently slipping, caused by fluctuations in the 



- 







K?j 




1 "1 

I 




r-^- 




-i^ 


— 


-B- 


^ 


^ 














' — ■ 




Fig. 715. 

power transmitted, the pulley rim may be cast with 
flanges, as shown in Fig. 716. See also Figs. 381 and 382, 
Art. 1500. 

2042. Arms of Pulleys. — The arms are usually of 
oval section, as shown in Fig. 717, A and 
B. It is customary to make the thickness 
\ the width. 

The arms are generally straight, as in 
Figs. 713 and 714, though curved arms are 
occasionally used. 

The number of arms to be used is largely 
a matter of judgment, but in practice, for 
all sizes of pulleys and engine band-wheels, 
under 10 feet in diameter, by far the 
greater number are made with 6 arms; 8 
or 10 arms are sometimes used for pulleys above 6 feet in 
diameter, and for very small sizes, 4 arms are suflicient. 




MACHINE DESIGN. 1335 

To calculate the width a of the arm, we may assume, as 

in gear-wheels, that the bending moment is p R^ and that, 

p R 
consequently, the bending moment for each arm is - — . 

The resisting moment of the section shown in Fig. 717 
is practically .04 a^ 5, where S is the allowable safe stress. 
If we allow 2,500 lb. per sq. in. as the safe stress, we have 



= 100 a\ or a= yJ. 



n ' 100/2* 

p =: T^— 7!j, the difference of the belt tension is usually 
not known exactly. Its maximum value can, however, 
easily be found, as follows : 

For single belts -^^" thick, an average tension of 320 lb. 
per sq. in. of section may be allowed. Then, 

T^ = ^\ X 320 = 70 lb. per inch of width of belt. 

Usually P is not more than -|- of T^, and its maximiun 
value may be taken at say 50 lb. per inch of width. For 
double belts, take P= 100 lb. per inch of width. 

Then, / = 50 ^ for single belts. 

/ = 100 B for double belts. ' 

Substituting these values of/ in the above equation. 



<2 = y — — , for smgle belts. 



In 



^^BR 



a =^ V • , for double belts. 



(277.) 



The taper in the width of the arms towards the rim may be 
made -|" per foot, and the thickness at the rim -|- the width. 

Example. — Calculate the size of the arms of a 6-arm pulley, 80 
inches in diameter, with a 6-inch face. 

Solution. — For single belt, we have, from formula 277, 



« = /^=//IKl5 = 2in..nearly. Ans. 
y 2« Y 2X6 ^ 

At the rim the width is 2" — (i X li) = 2" — f " = 1|". Ans. 
Thickness at center r= 2 X i = 1". Ans. 
Thickness at the rim = If" X i = -H"- Ans. 



1336 



MACHINE DESIGN. 



For very wide pulleys it is sometimes desirable to use two 
sets of arms, as shown in Fig. 718. 




Fig. 718. 

To calculate the size of the arms in this case, the pulley 
may be considered as made of two pulleys each of width ^ B. 
Then, find the dimensions of the arms as before, and multiply 
these dimensions by |/f = .8, nearly. 

2043. Hub of Pulley.— The thickness of the hub 
may be obtained from the following formula: 



w 



32 



+ ^" 



"S" 



(278.) 



The length of the hub may be, /= | ^ to B. 
The key-way may be calculated by the rules given in Art. 
1965, etc. 

2044. Loose Pulleys. — Pulleys which run loose on a 
shaft should have longer hubs than those keyed fast; the 
hubs may also be lined with brass bushings, if desired. 



MACHINE DESIGN. 



1337 



B 



mm^d^^m mm 



-Jk- I' 



w///////////m//y/ /mm 



Where a fast and a loose pulley are placed together on a 
shaft, as shown in Fig. 719, the length / of the hub of the 
loose pulley may be /= 1.2 
B^ and that of the fixed pul- 
ley I' = .^B. The thickness 
of the hub of the loose pul- 
ley may be less than that 
of the fast pulley on ac- 
count of its increased 
length. 

Some arrangement must 
be provided for oiling the 
loose pulley; generally one Fig. 7i9. 

or two oil holes are drilled through the hub. 

2045. For split pulleys the size of the bolts at rim and 
hub may be determined as follows: 

Let A = area of cross-section of rim ; 

A' = Sivea, of cross-section of hub along the line of 

division ; 
S = net section of bolt or bolts at rim ; 
S' = net section of bolt or bolts at hub. 

A 



Then, 5 



-\- ^ square inch. 

A' 
= — square inch. 



(279.) 



Example. — The hub of a split pulley is 4 inches long and 1^ inches 
thick. If the hub is held by 4 bolts, what should be the diameter of 
the bolts ? 

Solution. — A' = 4 X H = 5 sq. in. 

By formula 279, 

A' 

S' = -j- = f sq. in. = net section of 4 bolts. 

^-i-4: = ^^ sq. in. = net section of 1 bolt. 
Hence, from Table 43, Art. 1926, the diameter of bolt is f 
inch. Ans. 

2046. Wrougtit-iroii pulleys are coming into exten- 
sive use, and possess important advantages over those made 
of cast iron. They are lighter and stronger, and are free 



1338 



MACHINE DESIGN. 



from the initial stresses to which cast-iron pulleys are liable. 
Owing to the stresses due to centrifugal force, cast-iron 
pulleys can not be safely run at very high speeds; wrought- 
iron pulleys, however, may be used at almost any reason- 
able speed, because of the greater tenacity of the material of 
which they are made. Wrought-iron pulleys are usually split. 

2047. Maximum Speed of Cast-iron Pulleys. — 

"When a pulley rotates, each portion of the rim tends to fly 
outwards in the direction of a tangent to the rim. This 
tendency is due to the centrifugal force ^ so called. 

The centrifugal force on each element of the rim acts 
radially outwards, and the pulley under its action is anal- 
ogous to a section of a steam boiler under the pressure of 
steam. The rim tends to break at two sections which are 
at the opposite ends of a diameter. 

Table 51 gives values of the stress per square inch pro- 
duced in a pulley rim by centrifugal force for various 
velocities : 

TABLE 51. 



Velocity of rim in feet 

per second 

Velocity of rim in feet 

per minute 

Stress in rim in pounds 
per sq. in. : 

Cast iron 

Wrought iron 



60 


80 


100 


150 


3,600 


4,800 


6,000 


9,000 


351 


624 


975 


2,194 


378 


672 


1,050 


2,362 



200 



12,000 



3,900 
4,200 



To the stress given in the above table must be added the 
initial stresses due to contraction in cooling, and the 
stress caused by the belt pull. 

It is usually considered unsafe to run a cast-iron pulley, 
gear-wheel, or fly-wheel at a higher rim speed than 100 feet 
per second. Since the centrifugal force increases in direct 
proportion to the cross-section of the rim, it is evident that 
it is useless to try to provide against it By putting more 
material in the rim. 



MACHINE DESIGN. 



1339 



TABLE 52. 



B^ 


o 


Rim. 


Arm. 


Hub. 


Boss. 


c35 


A 


B 


C 


D 


E 


F 


G 


H 


I 


6 


4 


i 


A 


f 


tV 


3 


1 


i 


1 


X 




6 


.... 




< ( 


i ( 


3i 


i 


< 1 


( c 


1 1 




8 






( ( 


( ( 




1 ( 


1 1 


1 ( 


< < 




10 


. . . . 




( ( 


< ( 


4 


< ( 


( t 


1 ( 


(( 




12 


.... 


, . 


< ( 


( ( 


< < 


1 i 


< < 


1 < 


(( 


8 


4 
6 


i 


A 


If 


7 
T¥ 


3 

3i 


1 


i 


1 


i 




8 


A" 


i 


"ItV 


'i' 


^ 


. . 


. . 


. . 


5 , 




10 




, , 






5i 




. , 


, , 


. , 




12 


.... 


















lO 


4 
6 

8 


i 
^ 


i 


15 

Te 


1^ 


3 

3i 

4i 


i 


i 


1 


i 




10 






'ii 


"1" 


5i 


"i 


'i 


ii 


"i 




12 




















12 


4 
6 


3^^ 


i 


1 
If 




3i 
4 


i 


i 


1 


i 




8 










5 


'1 


'i 


ii 


'i 




10 


A + 


5 

T6 


"h' 


"f" 


5i 




. . 








12 










6i 










14 


4 


3 -4- 
3 2 ^ 


i 


H 


■^ 


3i 


i 


i 


1 


i 




6 










4^ 


f 


f 


n 


1 




8 


t\" 


5 


'li 


'i' 


5 












10 










6 












12 






'iH 


'if' 


6i 










16 


4 


s\ + 


i 


If 


t\ 


3i 


i 


i 


1 


i 




6 










4^ 


1 


f 


li 


1 




8 


t\ + 


5 

Te- 


'W^ 


"f" 


5 


. . 


. . 


. . 






10 










6 






, . 


, , 




12 


a" 


ll 

32 






6i 


'4 


. . 




. ^ 




16 






'ii* 


15 
T-g- 


8i 


I 


f 


if 




18 


4 
6 


3 


5 

Te- 


lA 


9 

Te- 


4 

41 


f 


f 


li 


f 




8 


3V 


ll 
33 


li' 


ll 
Te 


5i 


'f 




. . 


. . 




10 










6 






, . 


. , 




12 










n 


"i 


'f 


If 






16 


i" 


1 


'2i' 


n" 


8 








. . 




20 










9 










20 


4 
6 

8 


l\ + 


t\ 


n 


i 


4 

4^ 

5 


'f 


1 


li 


1 




10 


7 


11 
■3¥ 


'if' 


■f ■ 


6 






. . 






12 










7 








, , 




16 


9 
32 


7 
TH 


'2i 


n" 


8 


'■1 


'f 


If 






20 










10 


1 









1340 



MACHINE DESIGN. 



TABL/E BZ.— Continued. 



t 


(i5 
o 


Rim. 


Arm. 


Hub. 




Boss. 








A 


B 


C 


D 


E 


F 


G 


H 


I 


22 


4 
6 

8 


3 


T^ 


H 


f 


4 

4i 
5 




1 


li 


f 




10 


/^ + 


ii 


*if' 


"if 


6i 


.... 


, . 


. ^ 


• • 




12 


.... 


. . 






.... 


1 


f 


1| 


, , 




16 


.... 


, , 


, , 


^ 


8f 


1 










20 


3^^ + 


tV 


2i 


li 


11 


H 


'i 


ii 




24 


4 
6 

8 


-h 


ii 


1^ 


11 


4 

4f 
5i 




i 


li 


1 

• • 




10 


'i" 


1 


'ii 




7 


"i' 


'i 


ii 


• • 




12 


- • • ■ 




• • • . 


. • . • 


, , 


.... 










16 


.... 


, , 


. 


• • • 


9i 








• • 




20 


A 


1 5 
8T 


2f 


If 




H 


1 


ii 


^ ^ 




24 


.... 




.... 




ii" 








, , 


26 


4 
6 

8 


■h 


li 


Ill 


f 


5 
6 


"i' 




li 
ii 


1 




10 


i" 


i* 


'2* 


"*i* 


7 




, . 








12 






. « • • 




7i 


. . . 




^ ^ 


• • 




16 


.... 


. . 


. . • • 


.... 


10 


H 


'i 


H 


^ ^ 




20 


T^6 + 


if 


2H 


ItV 


lOi 








^ ^ 




24 










11 


.... 


. . 


. . 


, , 


28 


4 
6 

8 
10 


^ + 


M 


If 


f 


4i 

7 
7i 


.... 


f 


Ii 
If 


f 




12 


V+ 


1 


*2i' 


13 
T7 


8 




, , 


, , 


^ ^ 




16 


.... 








10 












20 


t\ 


ii 


.... 


.... 


11 


"ii' 


'i 


ii 


^ ^ 




24 


w 


i 


3i 


li 












30 


4 

6 

8 

10 


A + 


11 
IT 


n 


if 
.... 


4i 
6i 




1 
'i 


li 

if 


f 




12 


i" 


tV 


'2i' 




8 


'i" 


. , 


. . 


, , 




16 










8i 


. ■ • • 




, , 


, „ 




20 






. ■ • » 


.... 


IH 


li 


'i 


li 


, , 




24 


i" 


i 


3t^^ 


If 


13 


.... 








32 


4 
6 

8 
10 


i + 


1 


2i 


15 


4i 


.... 


f 


If 


« 




12 


5 


st" 


2tV 


"iiv 


8 


'ii' 


'i 


ii 


. , 




16 










9i 


.... 


, , 


, , 


, , 




20 










11 


u 




, , 


, , 




24 






.... 




13 








* 






MACHINE DESIGN. 



1341 



TABLE 52>,— Continued. 





0) 

o 


Rim. 


Arm. 


Hub. 


Boss. 




A 


B 


C 


D 


E 


F 


G 


H 


I 


34 


4 

6 

8 

10 


i + 


3 


2i 


if 




1 

i* 


f 


If 


f 




12 


ii" 


ii 


Vi' 


ixV' 


7f 


ii 


*i 


ii 






16 






.... 




9i 












20 


• • • • 




.... 


. • • • 


12 


ii 










24 


.... 


. . 


.... 


.... 


13 










36 


4 

6 

8 

10 


i+ 


f 


2x'^ 


15 


4i 
6f 


i 


f 


if 


f 




12 


xi" 


ii 
3^ 


.... 


.... 


-^f 


ii 


'i 


ii 






16 






3x^^ 


H 


lOi 


li 










20 


• • . • 




.... 




12 












24 


• . • • 


. . 


.... 


.... 


13i 


ii 


i' 


if 


'i 


40 


8 
12 


A 


15 
32 


3x^^ 


1 


6f 

7f 


1 


f 

1 


If 
li 


f 




16 


ii" 


V 


3f" 


ii" 


10 


li 










20 






.... 


.... 


Hi 


If 


i' 


if 


'i 




24 


.... 




.... 


.... 


15i 


li 








44 


8 
12 


A 


xV 


2i 


li 


6f 

8 


li 


1 


li 


1 




16 


11 

Z2 


V 


3'" 


ixi' 


10 


ii 


, , 


'] 






20 











12 


If 


1 


If 


I 




24 


.... 


, . 


3i 


.... 


15 


li 


, , 






48 


8 
12 


3'^ + 


xV 


2f 


If 


81 


li 


i 


li 


1 




16 
20 


1" 


i 


3i" 


ixi* 


10 
12 
15 


i| 


1 


if 


'i 




24 


.... 


;; 


.... 


> • • . 


ii 


;; 


;; 




54 


12 
16 
20 
24 


13 
33 


15 
33 

19 
33 


3 

3f" 


IxV 


Hi 
15" 


If 

li 

if 


1 

ii 


If 
2' 


i 


60 


12 
16 


11 
32 


i 


3x^. 


IxV' 


10 
Hi 


If 
li 


1 


If 


i 




20 


7 


V 


3ii' 


if" 


12i 


If 


ii 


2' 






24 






.... 


.... 


15 






. . 




66 


12 
16 


11 
32 


i 


3A 


Ix^^ 


10 

Hi 


li 


1 

li 


If 
2 


i 




20 


V" 


V 


ii" 


115 
■'■IS 


13i 


n 




. , 






24 


.... 








15 




. . 


. , 




72 


12 
16 


f 


x^^ 


H 


m 


lOi 
12i 


If 
If 


li 


2 


i 




20 


xi" 


13 
XS^ 


4"' 


2tV* 


13.^ 


ii 


. . 


. . 






24 










15 


2 









1342 



MACHINE DESIGN. 



2048. Examples of Belt Pulleys. — Table 52 gives 
the dimensions of a set of cast-iron belt pulleys ranging from 
6" to W in diameter, as made by a well-known manufac- 




Fig. 720. 

turing company. These pulleys are so designed that the 
number of patterns may be kept within reasonable limits, 
and at the same time have the dimensions correspond as 
nearly as possible with well-established rules. 

The letters over the columns of dimensions given, in the 
table correspond to the letters in Fig, 720. 

In all cases the number of arms is 6, and the arms increase 
in size towards the hub with a taper of ^' per foot. 

2049. Counterbalance. — Pulleys that run at high 
speeds must be carefully balanced, i. e., the center of gravity 
of the pulley must correspond with the center of the shaft, 
otherwise there will be a heavy stress on the shaft and 
bearings. Since it is seldom possible to make the pulley 



MACHINE DESIGN. 



1343 



exactly symmetrical, the difference in weight of the heavy 
side is compensated for by weights riveted to the inside of 
the rim on the light side. 

ROPE BELTING. 

2050. There is a growing tendency towards the sub- 
stitution of hemp and cotton ropes for belting and line shaf t- 




FlG. 721. 

ing as a means of transmitting power in large factories and 
shops. The advantages claimed for the rope-driving system 
are : 

1. Economy ; for a rope system is cheaper to install than 
either leather belting or shafting. 

2. In the rope system there is less loss of power by 
slipping. 

3. Flexibility ; that is, the ease with which the power is 
transmitted to any distance, and in any direction. 

2051. There are two systems of rope transmission in 
common use. In the first, the transmission is effected by 
several parallel independent ropes which pass around the fly- 
wheel of the engine and the pulley or pulleys to be driven. 



1344 



MACHINE DESIGN. 



Each rope is made quite taut at first, but stretches until it 
slips, after which it is re-spliced. A good example of a rope 
transmission of this character is shown in outline in Fig, 721. 
The fly-wheel D carries 35 parallel ropes which distribute 
power to the pulleys a^ b^ r, d^ e^ and y, located on the five 
floors of the mill. The ropes are distributed as follows : a, 
4 ropes ; b and c^ 6 ropes each ; d, e, and /", 7 ropes each. A 
secondary system of ropes drives the pulleys g^ h, k, and /. 



2052. In the second system of rope transmission, a 
single rope is carried around the pulley as many times as is 




Fig. 722. 

necessary to produce the required power, and the necessary 
tension is obtained by passing a loop of the rope around a 
weighted pulley. An example of this system is shown in 
Fig. 722. 

The rope is wrapped continuously around the fly-wheel 
D and the driven pulley E. From the last groove of E the 
rope is led over the idlers E and G, which are set at such an 
angle as to lead it back to the first groove in D. The weight 
W is attached to the pulley E which is movable along the 
rod H. The movement of the pulley E, therefore, takes up 
the stretch of the rope, and keeps it always at the same ten- 
sion. Rope pulleys may be attached to the shaft of the 



MACHINE DESIGN. 1345 

pulley E^ and the power received by E may thus be trans- 
mitted to any desired points. 

The first of the above systems of transmission is used 
chiefly in Europe; the second, in the United States. 

2053. Ropes. — The ropes used in rope transmission 
are either o£ hemp, manila, or cotton. Manila ropes are 
mostly used in this country. They are of three strands, 
hawser laid, and may be from \ inch to 2 inches in 
diameter. 

The weight of ordinary manila, or cotton, rope is about 
.3 D"^ pounds per foot of length, where D represents the 
diameter of the rope in inches. Letting w = the weight 
per foot of length, 

ze/ = .3i>^ (280.) 

The breaking strength of the rope varies from 7,000 to 
12,000 lb. per sq. in. of cross-section. The average value 
may be taken as 7,000 Z^^ when D is the diameter of rope. 

For a continuous transmission, it has been determined 

by experiment that the best results are obtained when the 

tension in the driving side of the rope is about -^ of the 

breaking strength. That is, 

7 000 D^ 
T = tension in tight side = ' — = 200 D^, 

So 

2054, Poiiver Transmitted by Ropes. — The ropes 
run in V-shaped grooves (see Fig. 724), and the coefficient 
of friction is, of course, greater than on a smooth surface. 
The coefficient for grooves with sides at an angle of 45° 

T 

may be taken at from .25 to .33. The ratio -^ will vary 

a 

from H to 3, depending upon the arc of contact and coefficient 
of friction. 

The horsepower that can be transmitted by a single rope 
running under favorable conditions is given by the formula 

D. 0. 111.— 25 



1346 



MACHINE DESIGN. 



The horsepower transmitted by ropes of different diam- 
eters running at different velocities may be calculated from 
formula 281, and plotted on cross-section paper. The 




O 10 2Q 30 4^ SO 60 70 80 90 100 
Velocity in Feet per Second . 

Fig. 723. 



1^ 130 140 ISO 



accompanying diagram (Fig. 723) shows the horsepowers 
transmitted by 1, 1^, 1|^, If, and 2 inch ropes for various 
velocities. The horizontal distances represent velocities in 
feet per second, and the vertical ordinates the horsepower 
transmitted by a single rope. 

The diagram shows that the maximum power is obtained 
at a speed of about 84 feet per second. For higher veloci- 
ties, the centrifugal force becomes so great that the power 
is decreased, and when the speed reaches 145 feet per 
second, the centrifugal force just balances the tension, so 
that no power at all is transmitted. Consequently, a rope 
should not run faster than about 5,000 feet per minute, and 
it is preferable, on the score of durability, to limit the 
velocity to 3,500 feet per minute. 

Example. — A rope fly-wheel is 26 feet in diameter, and makes 55 
revolutions per minute. The wheel is grooved for 35 turns of 1^-inch 
rope. What horsepower may be transmitted ? 



MACHINE DESIGN. 



1347 



V = 



Solution. — Velocity per second = 

26 X_0<55 _ 4,492 
60 ~ 60 

Then, from formula 281, 



= 74.9 feet. 



(^«»-S)=^«-"- 



825 V 107.2/ 825 V~"~' 107. 

the horsepower transmitted by one rope or turn. Then, 30.16 X 35 = 
1,055.6 = horsepower transmitted by the 35 ropes. Ans. 

Practically the same result may be obtained by referring 
to the diagram. It will be seen that for IJ-inch rope run- 
ning 74.9 feet per second, the horsepower per rope is slightly 
over 30; hence, the total horsepower is 30X35 = 1,050, 
nearly. 

Example. — How many times should a If -inch rope be wrapped 
around a grooved wheel in order to transmit 400 horsepower, the 
speed being 3,500 feet per minute ? 

3,500 



Solution. — 3,500 ft. per min. = 



60 



= 58^ ft. per sec. Referring 



to the diagram, a If-inch rope running at a speed of 58^ ft. per sec. 
transmits 36f horsepower. Hence, the number of turns should be 

gg- = 11, nearly. Ans. 

2055. Pulleys for rope g:eariiis: differ from ordinary 
pulleys in having a grooved rim. The sides of the groove 
are inclined at an angle which may vary from 30° to 60°, 




Fig. 724. 



The weight of the rope wedges it into the angle of the 
groove, and, therefore, the more acute the angle the greater 



1348 



MACHINE DESIGN. 



is the coefncient of friction, and likewise the wear of the 
rope. The general practice, at present, is to make the angle 
between the sides 45°. The grooves are made circular at the 
bottom, and are polished or smoothed to avoid wearing the 
rope. A section of what is known as the English form of 
grooved rim is shown in Fig. 724. The following proportions 
may be used: 

D = diameter of rope in inches. 



a=iB. 


e = ^D. 


d = liD. 


f=\D. 


c =\D. 


g = f-\-\ 


d = iD. 





A section of a grooved rim, in which the sides of the 
grooves are formed with circular arcs, is shown in Fig. 725. 

J 




Fig. 725. 



The proportions for this rim are as follows, using the diameter 
D of the rope as a unit: 



a = \D. 


e = \D + ^^" 


*=f^ + TV'- 


/=i^ + A" 


c^D. 


g=\D. 


d=1.6I). 


// = iZ) + T>/. 



r^ and r^ are to be found by trial ; they should be of such 
lengths as to make the curves drawn by them tangent to the 
required lines. 



MACHINE DESIGN. 



134& 



The long radius R is determined by drawing a line through 
the center of the rope at an angle of 22|-° with the horizontal, 
and producing it until it intersects a line drawn through the 
tops of the dividing ribs ; then, with this point of intersection 
as a center, draw the curve forming the side of the groove 
tangent to the circumference of the rope. 

The advantage claimed for this groove is that the rope 
will turn more freely in it, thus presenting new sets of 
fibers to the sides of the grooves, and increasing the life of 
the rope. 

2056. Guide pulleys, idlers, and tension pulleys 

do not have V grooves, but the rope rests upon the bottom 
of a circular groove, as shown in 
Fig. 726. 

2057. The diameter of a rope 
pulley should be at least 30 times 
the diameter of the rope. Good 
results are obtained when the 
diameters of pulleys and idlers on 
the driving side are 40 times, and 
those on the driven side 30 times 
the rope diameter. Idlers used 
simply to support a long span may 
have diameters as small as 18 rope 
diameters, without injuring the 
rope. 

When possible, the lower side of 
the rope should be the driving side, for in that case the rope 
embraces a greater portion of the circumference of the 
pulley, and increases the arc of contact. 

When the continuous system of rope transmission is used, 
the tension pulley should act on as large an amount of rope 
as possible. It is good practice to use a tension pulley and 
carriage for every 1,200 feet of rope, and have at least 10 
per cent, of the rope subjected directly to the tension. 

Aside from the grooved rim, rope pulleys are constructed 
the same as other pulleys. They may be cast solid, in halves 




Fig. 726. 



1350 MACHINE DESIGN. 

or in sections. The pulley grooves must be turned to 
exactly the same diameter; otherwise, the rope will be 
severely strained. 

WIRE-ROPE GEARING. 

2058. Telodynamic Transmission. — This name is 
applied to the method of transmitting power by means of 
wire ropes and pulleys. The method was introduced on the 
continent of Europe, in 1850, by C. F. Hirn, and has proven 
very successful and economical. Power may be transmitted 
great distances with very little loss. 

The telodynamic transmission is very simple. It consists 
of driving and driven pulleys connected by a wire rope run- 
ning at high velocity. When the distance is very great 
(sometimes several miles), relay pulleys are placed every 
400 to 500 feet. The driving pulley then drives the first 
relay pulley, which in turn drives the second, and so on, 
there being a separate rope for each relay. A single rope, 
however, may be used over a distance of 1,000 to 2,000 feet 
by supporting it by guide pulleys, as shown in Fig. 727. 
The guide pulleys should be not more than 500 feet apart. 






Fig. 727. 

The diameter of the guide pulley in the driving side of the 
rope should be equal to that of the driving pulley, while the 
diameter of the pulley supporting the slack side may be half 
as great. 

The pulleys are fixed to wood, iron, or masonry supports 
which are high enough to prevent the rope from dragging 
on the ground. 

2059. The wire rope used for transmitting power is 
usually composed of 6 strands twisted around a hemp core. 
Each strand is composed of either 7 wires or 6 wires, and a 



•MACHINE DESIGN. 1351 

central hemp core. The number of strands, and wires to 
the strand, may be varied at pleasure. 

Calling D the diameter of the rope, that is, the diameter 
of a ring that would just fit over the rope, and <^ the diam- 
eter of a single wire, we have for the ordinary rope of 42 
wires, 

i^=9^, or^=^. (282.) 

The weight of wire rope per foot equals 

w = lAd D\ nearly. (283.) 

Example. — What is the diameter of the wire composing a f-inch 
wire rope containing 42 wires, and what is its weight per foot ? 

D I- 
Solution. — ^=-^ = -^ = .07", nearly, or about No. 13 Brown & 

Sharpe's wire gauge. Ans. 

Weight per foot = 1.43 D'^ = 1.43 X {^Y = .56 lb. Ans. 

The total stress allowable in a rope of iron wires may be 
taken at about 25,000 lb. per sq. in. ; the stress in ropes of 
steel wire may be taken at 28,000 lb. per sq. in. 

In calculating the cross-section of a rope, the sum of the 
cross-sections of the individual wires must be taken ; thus, 
if n is the number of wires in the rope, the cross-section is 

^,d^ n, not-^Z>^ 
4 4 

2060. Tensions in a Suspended Rope. — A per- 
fectly flexible rope suspended between two points, as, for 
example, a wire rope suspended between two pulleys, hangs 
in a curve called the catenary. When, however, the de- 
flection is not very great, as is usually the case in wire-rope 
transmission, the curve is very nearly a true parabola. 

2061. In designing a wire-rope transmission, it is re- 
quired to know the distance the rope will hang below the 
points of suspension, i. e., the deflection of the rope, and 
the tension in the rope at the pulleys. 

In Fig. 728, let ^ C Bh^ 2i rope hanging from the points 
A and By which are supposed to be at the same elevation ; 



1352 



MACHINE DESIGN. 



then the tension a,t A or B is given by the formula 



(284.) 



H^ 




where w is the weight of the rope per foot of its length, 
r_ V 

-2a- 



c 

Fig. 728. 

^ = -| the distance in feet between the points of support, 
and // is the deflection in feet at the lowest point of the rope. 

Example. — The total distance between two rope pulleys is 400 ft., 
and the deflection at the center of the rope is 10 ft. Supposing the 
rope to weigh 1 lb. per ft., find the tension at the pulleys due to the 
weight of the rope. 

Solution.— « = 400 -4- 2 = 200 ; k = 10\ w = l. 

By formula 284, we have 

w«2 ^ ix (200)2 



T: 



w h 



+ 1 X 10 = 2,010 lb. Ans. 



2^ ' 20 

2062. If we solve formula 284 for h^ we obtain 



// = 



2w ^ A:w'' 2 * 



(285.) 



This formula may be used to find the deflection when the 
tension, weight of rope, and span are known. 

2063. Stresses in a Wire Rope. — A wire rope when 
transmitting power is subjected to three different stresses: 

1. The stress due to the direct longitudinal tension 
which depends upon the span, power transmitted, and 
weight of rope. 

2. A stress caused by bending the wire around the 
convex portion of the pulley or sheave. 

3. Stress due to centrifugal force. 

The values of these stresses are expressed by the following 
formulas: The stress per square inch due to direct 
tension is 



wa^ 



MACHINE DESIGN. 1353 



-\-w]i 



5, = l^— = ^"(; + f-'l (286.) 

— d'^n 
4 

The stress per square inch due to bending around the 
pulley equals 

5. = ^, (287.) 

where E^ is the coefficient of elasticity of the material com- 
posing the wire (see Art. 1352), <3r= diameter of wire 
composing the rope, and R = radius of pulley in inches. 
The stress per square inch due to centrifugal force is 

where v = velocity of rope in feet per second, and g = 32.16. 
The total stress per square inch is the sum of these 
stresses, and is equal to 

^-^, + ^, + ^,_ r^^;^/^ ^2R^7rd'ng' <^^^') 

This total stress »S should not exceed 

25,000 lb. per sq. in. for wrought iron; 
28,000 lb. per sq. in. for steel. 

2064, Ratio of Tensions. — With long-rope trans- 
missions the arc of contact does not vary much from 180°, 
and the coefficient of friction may be taken as about .22. 
With these conditions the tension T^ on the driving side is 
about twice the tension T^ on the slack side; i. e., T^ = ^T^^ 
therefore, the driving force P = ^i "~ ^-i — -^i ~ i -^» 

The horsepower transmitted is 



550 33,000' 

One or two examples will serve to illustrate the use of 
the above equations and formulas. 



1354 MACHINE DESIGN. 

Example. — Power is transmitted by an iron wire rope containing 42 
wires No. 15 B. & S. wire gauge, or .057 inch in diameter. The pulleys 
are 11 feet in diameter, and the rope runs at a speed of 4,800 feet per 
minute. The distance between pulleys is 400 feet. Find the horse- 
power transmitted, and the deflections of both tight and driving sides 
of rope. 

Solution. — The. stress due to bending is 

c — -^lA: 

For iron Ex is about 25,000,000. Hence, 

^ 25,000,000 X. 057 ,^ ^^r. ik 

^^ rn7T2 ^ 10,795 lb. per sq. in. 

2X ^ 

Diameter of rope is, by formula 282, 

Z> = 9^=9x.057 = .513", 
Weight of rope per foot is, by formula 283, 

w = 1.43 D'^ = 1.43 X .5132 = .3764 lb. 
Stress due to centrifugal force is 

n^^A /4,800\2 

•^^ = ^f7^^ = 3.1416 X. 057^X42X32.16 = ^^^ ^^' ?"' "^- '''' 

St can not be calculated by formula 286, as the value of ^ is not 
known ; but from formula 289, 

St=S-{Sb + Sc), 
and since 5 must not exceed 25,000 for wrought iron, 

St = 25,000 - (10,795 + 699) = 13,506 lb. per sq. in. 

Ti = maximum tension on driving side = 'jd^nSt — 

.7854 X .0572 x 42 X 13,506 = 1,448 lb. 
Tu = iTi = 724 lb. ; F = driving force = T^ = 724 lb. 

zr u ^^ 724x4,800 ,_ _ . 

Zf^ horsepower =.-33^^ 33,000 =^^^'^' ^^"- 

For the deflection, we may use formula 285, 



3w r 4^2 2 

The deflection of the driving side is, taking 'w as -f, 
_ 1,448 /T448^ {260Y „ ,^ . 

The deflection of the driven side is 

J72^ ^/"^724^ (200? ,^ , .^ . 



MACHINE DESIGN. 1355 

In order to solve the converse problem, that is, find the 
necessary size of a wire rope to safely transmit a given 
horsepower, some assumptions must be made. In the first 
place, the stress due to bending can not be found directly, 
since the diameter of the wire is unknown. We may, how- 

ever, assume a ratio for — ^, and after finding the size of the 

wire required, the radius of the pulley is at once known. 

This ratio of -j varies from GOO to 1,400. When —f = 850 for 
a a 

iron wire, the pulley diameter will be smaller than for any 
other ratio. The ratio should, however, be as great as the 
conditions will allow, for the larger the pulley for a given 
diameter of wire, the greater is the durability of the 
rope. 

If the value of 5^, the stress due to the centrifugal force, 
be taken into account, it will complicate the solution very 
much; in fact, the only method of solution will be a cut- 
and-try method. As S^ is small in comparison with S^ or Sf_ 
for reasonable values of v^ it may be neglected, and the diam- 
eter of the wire calculated as in the next example. If 
greater exactness is desired, substitute this value of d in for- 
mula 288 ; calculate the value of S^^ and then re-calculate 
the value of d. 

Example. — Required, to find the diameter of the wires in a steel 
wire rope transmitting 200 horsepower at a velocity of 5,100 feet per 
minute ; also, required, the diameter of the pulley and rope. The rope 
is to contain 42 wires. 

Solution. — The driving force equals 

„ _ 33,000 H _ 33,000 X 200 . ._ . ,, 
F - j^~ - • ,^^ = 1,294 lb. 

The tension on driving side equals 

Ti = 2 P = 2 X 1,294 = 2,588 lb. 

/? 
Assume the ratio -^ = 900 ; E^ for steel = 30,000,000. 



Then, the stress due to bending is 
S - -5*1^ _ 30,000,000 _ 
^ 2 7? ~ 2 X 900 ~ 

Hence, . St = 28,000 - 16,667 = 11,333 lb. per sq. in. 



C £ld 30,000,000 ^nr.nr>^^. 



1356 MACHINE DESIGN. 

Cross-section of wires = 

1 J'> ^1 2,588 rtnoo 

ir^^;. = -^= ^^^333. = . 2283 sq.m. 



d = jJ ^.:^'!"^"' ., = .0832 in. 



■ 2283 
7854 X 42 
The diameter of the rope is, therefore, 

D = 9d = 9x .0832 = .7488, or say f in. 
Assuming the diameter of rope to be f" = .75", as found, the weight 
per foot is by formula 283, 

w = 1.43 £>'' = 1.43 X .75-^ = .8044 lb. 

By formula 282, d=~ = '-^ = .0833. 

Hence, by formula 288, 

/ 5,100 Y 
c _ 4 X. 8044x1 60 ) _ 

" ~ 8.1416 X .0833'^ X 42 X 32.16 ~ 
And, St = 28,000 - (16,667 + 790) = 10,543 lb. per sq. in. 
Cross-section of wires = 

i.^-^;.=-g=^ = .2455sq.in., 



and ^=i/ •^'*""' = .0862. Ans. 

r .7r ' 



.2455 

r85rx42 

The diameter of the rope is, therefore, 

.0862x9 = .7758", say f ", 
the same value as obtained when So was neglected. Ans. 

Since ~ = 900, 7? = 900 <^= 900 X .0862 = 77.58", and the diameter of 
a 

the pulley ==2 X '^"7.58 = 155" = 12 ft. 11". Ans. 



2065. Wire ropes are used also for hoisting and hauling 
loads in and about mines. The calculation of the size of a 
wire rope for any purpose is similar to that given above. If 
the rope is subjected to a straight pull, and is not bent around 
a sheave or drum, the cross-section may be found directly 
from the load and allowable safe stress; that is, cross- 
total load on rope 

section = — ^ ^ — . 

sate stress per sq. m. 

When the rope is bent around a pulley or drum, the stress 

E d 
due to bending, that is, S^ = -^7^, must be subtracted from 

the total safe stress. 



MACHINE DESIGN. 



1357 



Wire ropes used for hoisting are often made with 6 strands 
of 19 wires each, wound around a central hemp core, making 
in all G X 10 = 114 wires. This rope is more flexible than 
the regular transmission rope, and is, therefore, injured less 
in passing over small pulleys ; 
but it will not stand as much 
wear when dragged over rough 
surfaces, as the wires of which it 
is composed are so much smaller. 

2066. ^Wire-Rope Pul- 
leys. — The pulleys, or sheaves, 
used in wire-rope transmission 
are made of cast iron with a 
groove lined with rubber, gutta- 
percha, leather, or other similar 
substance, on which the rope 
runs. The grooves are made so 
wide that the rope rests on the 
rounded bottom instead of being 
wedged against the sides, as in the case of hemp or cotton 
rope. 

The proportions of the pulley rim are shown in Fig. 729. 

They are as follows: 

<af= diameter of rope; 

c = '^d. g = \d-\-^'. 

The arms may be cross-shaped or oval ; the latter form is 
preferable, as it offers less resistance to the air when the 
pulley is run at high speed. The size of the arms corre- 
sponds to those of belt pulleys transmitting the same force. 
For the number of arms Reuleaux gives the formula 




Fig. 729. 



n = ^,-\- 



R 



40 Z>' 



(290.) 



where R = radius of pulley in inches and D = diameter of 



1358 



MACHINE DESIGN. 



rope. The diameter of the pulley is fixed by the diameter 
of the rope and the number of wires in a strand. For a 
rope with seven wires to the strand, the diameter of the 
pulley should not be less than 150 times the diameter of the 
rope; and for a rope with 19 wires to the strand, the propor- 
tion should not be less than 90 to 1. 

Table 53 gives the breaking strength and the power trans- 
mitted by various sizes of ropes, as determined by practical 
experience: 

TABLE 53. 



POTVER TRANSMITTED BY If^TIRE ROPES (4=3 WIRES). 



Diam- 


Diam- 


Revolu- 


Breaking 


Horse- 


Velocity of 


eter of 


eter of 


tions 


Stress of 


power 


Rope in 


Ropes, 


Pulleys, 


per 


Rope per 


Trans- 


Feet per 


Inches. 


Feet. 


Minute. 


Pound. 


mitted. 


Second. 


7 


5 


100 


4,260 


8.6 


26 


1 5 
32 


6 


100 


5,660 


13.4 


31 


i 


7 


100 


8,200 


21.1 


36 


f 


8 


100 


11,600 


27.5 


42 


1 


8 . 


120 


11,600 


33.0 


50 


1 


9 


100 


11,600 


51.9 


47 


f 


9 


120 


11,600 


62.2 


56 


11 
16 


^ 10 


100 


15,200 


73.0 


52 


H 


10 


120 


15,200 


87.6 


62 


11 
16 


10 


140 


15,200 


102.2 


73 


« 


12 


100 


15,200 


116.7 


63 


f 


12 


120 


17,600 


148.9 


75 


f 


12 


140 


17,600 


173.7 


87 


i 


14 


100 


17,600 


185.0 


73 


i 


14 


120 


17,600 


222.0 


87 


f 


15 


120 


17,600 


300.0 


94 



Note. — The student may obtain much information concerning 
wire ropes from the trade catalogue of John A. Roebling's Sons Co., 
Trenton, N. J. 



MACHINE DESIGN. 



1359 



CHAINS. 

2067. Chains may be used as simple fastenings or as 
belts for transmitting power. The ordinary, or open-link 
and the stud-link round iron chains are shown in Fig. 730. 
The links are made from round iron bars which are cut ofi. 
at the proper length, bent, and welded. The links should be 
made as small as possible, both on account of strength and 
flexibility. Ordinary chain proportions are shown on the 
figure. They are as follows: 

d = diameter of iron ; 

a = 4:id to 5d; 



For open link 



For stud link -< 



b—^d. 



a = 5 d to 6 d; 
b = 3idto 3f^; 
c = .Qd; 
e=.7d. 



Link chains which are used merely to support loads, as in 
suspension bridges, etc., have the links from 3 to 9 feet or 
more in length. As such 
chains do not belong properly 
to the subject of Machine 
Design, they will not be con- 
sidered here. 

2068. Strength of 
Chains. — The strength of a 
chain is less than that of the 
iron composing it, on account 
of the weld, and also because 
of the presence of bending 
action. 

Formulas 130 and 131, 
Art. 1421, may be used to ^'^- ^^o. 

find the safe load in ordinary cases. For crane chains which 
require a large factor of safety, Towne gives the following 
as the safe load: 




P= 3.3^' tons = 6,600 ^' lb. 



(291.) 



1360 



MACHINE DESIGN. 



The weight of chains (open and stud link) may vary from 
9 d'^ to 9^ d^ pounds per foot. 

2069. Chain Drums. — When a chain must be coiled, 
as in the case of cranes and derricks, a grooved drum may 
be used. The groove passes spirally around the drum, and 
is just wide enough to receive the edge of a link of the 
chain. The drum may have a diameter of from 24 ^ to 30 d 
or more ; the length should be such that the total amount of 
chain may be coiled on in one layer ; because, if one layer is 
wound over another, the chain is injured. 

Instead of a drum, a wheel or sheave with pockets may 
be used. Such a wheel requires less space than the drum, 
and injures the chain less. A form of chain wheel largely 






Section CD 




Section AB 

Fig. 731. 

used for transmitting power, especially on cranes, chain 
blocks, etc., is shown in Fig. 731. The rim of the wheel is 
grooved for the links, and pockets are provided into which 
the links that lie parallel with the axis of the wheel rest. 

The pitch of the pockets must, of course, be the same as 
the pitch of the links. 



MACHINE DESIGN. 



1361 



2070. Flat-link ctiains are used for driving ma- 
chinery where very heavy resistances are to be overcome, 
as, for example, in wire-drawing machines, cranes, and 
dredging machines. 

When the chain merely supports a load, it may have the 
form shown in Fig. 
732. It consists of 
flat plates which are 
connected by pins. 
The pins are evi- 
dently in shear, and 
the plates are in 
direct tension. 
Since each of the 
two parallel plates 
carries one-half the 
load, it should have fig. 732. 

one-half the thickness of the single plate to which it is 
pmned. 

The links may be of any length desired, but the shortest 
convenient length is about I =.Zd. The cross-section = ab 

P 

— -^, where P is the load and S^ the safe stress in tension. 

Taking both P and S^ in tons, we may assume 5^ = 5 
tons. 

Then, ab — .%P. 

The shearing section of a pin is 2 X i^r^a^'' = ^7r<2^^ There- 
fore, 

P 




i^d'z 


~s: 










where 5^ = safe shearing stress 


. 










Assuming 5^ = 4 tons, we have 










d=V 


^p 

2rr* 










The following proportions 


may 


be used 


in 


ord 


mary 


cases : 












b=\d c=^ 


d. e 


= ^d 








D. 0. IIL—.4(J 













13G2 



MACHINE DESIGN. 



When the links are short, the width b may be the same 
throughout. The pin connecting the plates may be riveted 
over or secured by a washer and split pin. 

A flat-link chain may be used for transmitting powers 
somewhat after the fashion of a belt. The chain passes 
over wheels provided with teeth which engage with the links 
of the chain. Such a wheel is known as a sprocket wheel. 
Examples of chains used in this way are met with in 
agricultural machinery, bicycles, coal-mining machines, 
dredges, etc. 

The flat-link gearing chain, Fig. 733, consists of two series 
of flat links which are kept some distance apart by the pins 



K- I — ^ 



Y^^d^ 





Fig. 733. 



[3^MlJ^^^ ^^4^J^^ 



which connect them. These pins engage with the teeth of 
the sprocket wheel ; they are enlarged between the series of 
plates so as to form a shoulder to prevent the plates from 
slipping, and also to give a greater wearing surface. 

Let 11 = number of plates on one side of chain ; 
a and d = thickness and breadth of plates, respectively; 
d= diameter of ends of pin; 
d'= diameter of center of pin; 
A = length of enlarged part of pin ; 
/ = length of link between centers of pins; 
/^= total load on chain in pounds. 

All dimensions in inches. 



MACHINE DESIGN. 1363 

Then, the following formulas and proportions are gener- 
ally used: 

^ = .13y^; /?= 1.7^+. 5; 



/-' 



b — %.hd\ a — — —7; 

' n -\- r 

d'=1.2d; e=.%bd. 

Example. — Calculate the dimensions of a plate-link gearing chain 
for a working load of 8,000 lb. 

Solution. — Number of plates on one side = 



/z = .13 -f P = .13 f 8,000 = 2.6, say 3; 

d =.0115 i/ -^ = .0115y -^ = .78, say f"; 

y fl y O 

^' = 1.2 //= 1.2 X f = .9", say ft"; 

b =2.5^=2.5xf = 11"; 

k = 1.7rtr4-. 5 = 1.7 Xf + .5 = 1.775"; 

/ =2.9^=2.9X1 = 2.175"; 

.85./ _ .85 Xf _ .0. sa„i.s 
"" -ITT?- 3 + 2.175-'^^ ,sayi ; 
e =.85^=.85xf =.64". 

Various other forms of gearing chains are in common use. 
In designing machinery requiring the use of such chains it 
is customary to use some standard size made by a company 
engaged in the manufacture of chains. 

It would be well for the student to send for the catalogue 
of some such company, and observe the proportions and 
sizes there adopted. Much good information may be 
obtained from the catalogue of The Link Belt Engineering 
Company, Chicago, 111. 

2071. Hooks. — Chains used on cranes and derricks 
must be provided with hooks for connecting to the load 
to be raised. The design of a crane hook must be made 
with care, since a break may cause loss of property and 
life. 

A hook may be treated theoretically, as follows: The 



1364 



MACHINE DESIGN. 



maximum stress comes on the section m n of the hook. 

See Fig. 734. Let the center of 
gravity of this section be a distance 
a from the line of action of the load, 
and a distance c from the inside edge 
of the hook. 

Let S^ = the stress in the section 
due to direct tension; 

Sf, = the stress due to bend- 
ing; 

A = area of section; 

/ = moment of inertia of sec- 
tion. 

p 
Then, for direct tension, P= A Sf, or 5^ = ^. 

For bending, letting Sf, = safe bending stress, the moment 




Pa = ^ (see Art. 1398), or 5, 



Pac 
I ' 



Now, Si,-\- Sf^ equals total stress, must not exceed a safe 
value, say about 12,000 lb., or 6 tons per sq. in. Then, if 
we take P in tons, 

S, + S, = 6 tons = p(l+^y 

Suppose, for example, we assume the section of the hook 
to be a rectangle, of length d and width | d, and let the 
distance a^ Fig. 734, equal the length d of the section; 
what should be the dimensions of section for a load of 3 
tons ? 

We have 



21 



or b' 

Hence, b = y'^ = 2.29, say 2^"; 
and 1^=1.53, say 1J|". 



MACHINE DESIGN. 



1365 



The proportions shown in Fig. 735 are those used by a 
large crane manufacturing com- 
pany. They are based on the 
diameter of the iron rod of which 
the hook is made. This diameter 
may be obtained by the following 
formula: 





3^d. *J 



Fig. 735. Fig. 736. 

d = \/P^ where P is the load in tons. 

The hook may have an eye, as shown in the figure, or a 
neck for a swivel, as shown at A. 



PIPE FLANGES. 

2072. The ends of cast-iron pipes, elbows, etc., are 
often connected by means of flanges and bolts. 

Fig. 736 shows the method of flanging and bolting the 
ends of two cast-iron pipes. The dimensions of the flanges 
for the various sizes of pipes are given in the following 
table : 



1366 



MACHINE DESIGN. 
TABLE 54. 



STANDARD PIPE FLANGES. 

n = number of bolts. 



a 


b 


c 


ef 


n 


e 


/ 


g 


2.0 


.409 


5 

8 


2.0 


4 


f 


4.75 


6.0 


2.5 


.429 


f 


2.25 


4 


1 1 

16 


5.25 


7.0 


3.0 


.448 


5 

8 


2.5 


4 


f 


6.0 


7.5 


3.5 


.466 


f 


2.5 


4 


1 3 
1 6 


6.5 


8.5 


4.0 


.486 


f 


2.75 


4 


W 


7.25 


9.0 


4.5 


.498 


f 


3.0 


8 


1 5 
16 


7.75 


9.25 


5 


.525 


f 


3.0 


8 


1 5 
16 


8.5 


10.0 


6 


.563 


f 


3.0 


8 


1 


9.625 


11.0 


7 


.600 


f 


3.25 


8 


ItV 


10.75 


12.5 


8 


.639 


f 


3.5 


8 


H 


11.75 


13.5 


9 


.678 


f 


3.5 


12 


li 


13.0 


15.0 


10 


.713 


i 


3.625 


12 


lA 


14.25 


16.0^ 


12 


.790 


i 


3.75 


12 


li 


16.5 


19.0 


14 


.864 


1 


4.25 


12 


If 


18.75 


21.0 


15 


.904 


1 


4.25 


16 


If 


20.0 


22.25 


16 


.946 


1 


4.25 


16 


-•-16 


21.25 


23.5 


18 


1.020 


li 


4.75 


16 


1 9 
-•-16 


22.75 


25.0 


20 


1.090 


H 


5.0 


20 


111 
■^16 


25.0 


27.5 


22 


1.180 


li 


h.h 


20 


HI 


27.25 


29.5 


24 


1.250 


H 


5.5 


20 


l| 


29.5 


32.0 


26 


1.300 


H 


5.75 


24 


2 


31.75 


34.25 


28 


1.380 


H 


6.0 


28 


^16 


34.0 


36.5 


30 


1.480 


If 


6.25 


28 


n 


36.0 


38.75 


36 


1.710 


If 


6.5 


32 


2f 


42.75 


45.75 


42 


1.870 


li 


7.25 


36 


2| 


49.5 


52.75 


48 


2.170 


li- 


7.75 


44 


2f 


56.0 


59.5 



2073. The larger sizes of wrought-iron pipe, especially 
for high pressures, are best connected by means of flanges. 
Fig. 737 shows three styles of fastening these flanges to the 
pipe. 



MACHINE DESIGN. 



1367 



^ is a cast-iron flange screwed to the end of the pipe 
by means of the standard pipe thread. It is much used, 
especially for sizes of pipe below 6". 

^ is a wrought-iron or steel flange which is screwed to 
the end of the pipe, and the end of the pipe is then riveted 
over as shown at a. 

6' is a forged or rolled steel flange that is used for large 
pipes and high pressures. The pipe is fastened to the flange 




Fig. 737. 

by rivets as shown at ^, and the end is also calked or riveted 
over at d so as to form a tight joint. 

The diameters of flanges and bolt circles, and number and 
sizes of bolts for wrought-iron pipe flanges, should be the 
same as given in Table 54; 
for, then, the flange for a 
pipe of a given size, whether 
cast or wrought iron, will fit 
any other pipe, valve, or fit- 
ting of the same size. 

The thickness of the 
flanges may also be the 
same, although wrought- 
iron and steel flanges are 
usually made somewhat 
thinner. 

2074. Gaskets. — In 

order to secure a tight joint, 
a thin strip of some soft 
material is placed between 
the flanges as shown in Fig 




38. 

material 



most 



1368 MACHINE DESIGN. 

commonly used is either sheet rubber or paper, cut to the 

size of the flange. 

For steam pressures above 100 
pounds per square inch, however, 
these materials are injured by the 
heat and often make trouble. To 
overcome these difficulties gaskets are 
sometimes made of lead or copper. 
Fig. 739 shows a gasket made of thin 
Fig. 739. sheets of corrugated copper, as shown 

in the section. 




A SERIES 

OF 

QUESTIONS AND EXAMPLES 

Relating to the Subjects 
Treated of in This Volume. 



It will be noticed that the questions and examples con- 
tained in the following pages are divided into sections 
corresponding to the sections of the text of the preceding 
pages, and that each section has a headline which is the 
same as the headline of the section to which the questions 
refer. No attempt should be made to answer any questions 
or to work any examples until the corresponding part of 
the text has been carefully studied. 



ELEMENTARY MECHANICS. 

(ARTS. 828-966.) 



EXAMINATION QUESTIONS. 

(355) A ball, thrown horizontally by the hand, has a 
velocity of 500 ft. per second. If the ground is level, and 
the distance from the ground to the hand at the instant the 
ball leaves the hand is 5 ft. 6 in., how far will the ball go 
before striking the ground ? Ans. 292. 42 ft. 

(356) An engine fly-wheel, 80 in. in diameter, makes 160 
revolutions per minute; what is the velocity in feet per sec- 
ond of a point on the rim ? Ans. 55.85 ft. per sec. 

(357) In the last example, through how many degrees, 
minutes and seconds will a point on the rim turn in one- 
seventh of a second ? Ans. 137^ 8' 34f ". 

(358) The fly-wheel of an engine drives a pulley, rigidly 
connected to a drum on which an elevator rope winds. The 
fly-wheel is 4 ft. in diameter and makes 54 revolutions per 
minute; the diameter of the pulley is 36 in., and of the 
drum, 18 in. ; (a) how long will it take the elevator to reach 
the top of a building 100 ft. high ? (3) If required to reach 
it in 30 seconds, how many revolutions should the fly-wheel 
make per minute ? » j (a) 17.68 sec. 

■ ( (^) 31.83 R. P. M. 

(359) Define and give an example of uniform motion ; of 
variable motion. 

(360) Define force. Name five different kinds of forces. 

(361) Explain what you understand by inertia. 

(362) Is inertia a force ? If so, why ? 

(363) What is weight ? How is it measured ? 

(364) In order that the effect of a force upon a body may 
be compared with that of another force acting on the same 
body, what three conditions must be fulfilled ? 

For notice of the copyright, see paj^e immediately following the title page. 



568 ELEMENTARY MECHANICS. 

(365) What is motion ? What is rest ? 

(366) Can a body be in motion with respect to one body 
and at rest with respect to another ? Give examples. 

(367) Where will a body weigh the more, on the top of a 
high mountain or at the bottom of a deep valley, the bottom 
of the valley being as far below sea-level as the top of the 
mountain is above ? 

(368) If the top of a mountain is 31,680 feet above 
sea-level, what would a body weighing 20,000 lb. at sea- 
level weigh at the top of the mountain ? Take the radius of 
the earth at sea-level as 3,960 miles. Ans. 19,939 lb. S^ oz. 

(369) If the body in the last example had been dropped 
in a hole just big enough to let it fall to the bottom, and the 
bottom of the hole was 2 miles below sea-level, how much 
would it have weighed at the bottom of the hole ? 

Ans. 19,989 lb. 14.4 oz. 

(370) State the three laws of motion. 

(371) What is acceleration ? 

(372) What do you understand by initial velocity ? 

(373) Why is it dangerous to jump from a moving train ? 

(374) Why is it that a man cannot lift himself by pulling 
his boot straps ? 

(375) Explain how forces are represented by lines. 

(376) What is meant by the expression, "the resultant 
of several forces "? 

(377) If a line 5 in. long represents a force of 20 lb., {a) 
how long must the line be to represent a force of 1 lb. ? (d) 
Of 6i lb. ? 

(378) What do you understand by the components of a 
force ? 

(379) If a body be acted upon by two equal forces, one 
due east and the other due south, in what direction will the 
body move ? What is the direction of the. resultant of the 
two forces ? 

(380) Find the point of suspension of a rectangular cast 
iron lever 4 ft. 6 in. long, 2 in. deep and f in. thick, having 



ELEMENTARY MECHANICS. 569 

weights of 47 lb. and 71 lb. hung from each end, in order 
that there may be equilibrium. Take the weight of a cubic 
inch of cast iron as .261 lb. 

Suggestion. — First find the weight of the lever; then consider this 
weight to be concentrated at the center of gravity of the lever, and 
combine it with the other two weights in the same manner as though 
there were three weights. 

A j vShort arm ==22.342 in. 
( Long arm = 31.658 in. 

Solve the two following examples by the inethod or tri- 
angle of forces, and parallelogram of forces, and mark the 
direction of the resultant : 

(381) Two forces act upon a body at a common point; 
one with a force of 75 lb., and the other with a force of 40 
lb.; if the angle between them is 60°, and both forces act 
towards the body, what is the value of the resultant ? 

Ans. 101.12 lb. 

(382) In the last example, if one force (the one of 75 lb.) 
acts away from the body, and the other towards it, what is 
the resultant ? Ans. 65 lb. 

(383) If two forces of 27 lb. and 46 lb., respectively, act 
in exactly opposite directions upon a body, what is the 
resultant ? 

(384) The entire solar system is moving through space 
at the rate of 18 miles per second; (a) what is its velocity 
in miles per hour ? (l?) How far will it go in one day ? 

(385) Two bodies, starting from the same point, mdVe 
in opposite directions, one at the rate of 11 ft. per second, 
and the other at the rate of 15 miles per hour; (a) what 
will be the distance between them at the end of 8 minutes ? 
(d) How long before they will be 825 ft. apart ? 

Ans. 1 ('') 3 "^"es. 

( (l?) 25 seconds. 

(386) If six forces act towards the center of gravity of a 
body at angles of 30°, 45°, 135°, 210°, 225°, and 300°, whose 
magnitudes are 75, 47, 61, 32, 53, and 98 pounds, respectively, 
what is the value of their resultant ? Solve by method of 
polygon of forces. Ans. 45f lb. 



570 ELEMENTARY MECHANICS. 

(387) Suppose that the velocity of a steamboat in still 
water is 10 miles per hour, and that it is placed in a river 
flowing 4 miles per hour; also, that a man is walking the 
deck from stern to bow at the rate of 3 miles per hour, (a) 
What is the velocity of the boat when headed up stream ? 
[If) When headed down stream ? (c) Of the man in each case ? 

(388) A peg in the wall is pulled by two strings, one 
with a force of 21 lb., at an angle to the vertical of 45°, and 
the other with a force of 28 lb., at an angle of 60°; what is 
the value and direction of the resultant when the forces are 
on opposite sides of the vertical line ? Use the method of 
parallelogram of forces. Ans. 30.34 lb. 

(389) A force of 87 lb. acts at an angle of 23° to the hori- 
zontal; what are its horizontal and vertical components? 
Find, first, by the method of triangle of forces, and, second, 
by trigonometry. » j 80.084 lb. 

^^' (33.994 1b. 

(390) A weight of 325 pounds rests upon a smooth 
inclined plane, as shown in Fig. 119, Art. 884. If the 
angle of the plane is 15°, (a) what is the perpendicular pres- 
sure against it ? (d) What force would it be necessary to 
exert parallel to the plane, to keep it from sliding down- 
wards, there being no friction ? Solve by trigonometry, and 
also by the method of the triangle of forces. 

^^g j(^) 313.93 lb. 
\ {b) 84.12 lb. 

(391) If the weight of a body is 125 lb., what is its mass ? 

(392) If the mass of a body is 53.7, what is its weight ? 

(393) {a) Is the mass of a body always the same ? {b) 
If the mass of a body on the earth's surface is 25, what 
would be its mass at the center of the earth ? {c) On the 
surface of the moon ? 

(394) A body on the earth's surface weighs 141 lb. ; [a) 
at what point above the surface will it weigh 100 lb. ? {b) At 
what point below the surface ? Take the earth's radius as 
4,000 miles. ^^^ j {a) 749.736 miles. 

^^' \ \b) 1,163.12 miles. 



ELEMENTARY MECHANICS. 571 

(395) If a body were dropped from a balloon 1 mile above 
the earth's surface, (a) how long a time would it require to 
fall to the earth ? (d) What would be its velocity when it 
struck ? ^^^ j (a) 18.12 seconds. 

I {d) 582. 7G ft. per sec. 
(o96) In the last example, if the body weighed 160 lb., 
what would be the kinetic energy on striking the earth ? 

Ans. 844,799 ft. -lb. 

(397) If a cannon ball were fired vertically upward with 
a velocity of 2,360 ft. per second, (a) how high would it go ? 
(d) How long would it take to return to the earth ? 

Ans i^"^^ 16.4 miles. 

■ I (^) 2 min. 26.77 sec. 

(398) The earth turns round once in 24 hours. If it -w^re 
a perfect sphere 8,000 miles in diameter, how far would a 
point on the equator travel in one minute ? 

(399) If a projectile weighing 400 pounds be fired from 
a cannon with a velocity of 1,875 ft. per second, at a target 
6 ft. distant, (a) what will be its kinetic energy, on striking 
the target, in foot-pounds ? (l?) In foot-tons ? (c) If it pene- 
trates but 6 in., what will be its striking force ? 

{ (a) 21,863,339.55 ft.-lb. 
Ans. j (d) 10,931.67 ft. -tons. 
( (c) 43,726,679 lb. 

(400) If the acceleration due to gravity were 20 ft. per 
second, instead of 32.16 ft. per second, how much longer 
would it take a body to fall to the earth from a height of 200 
ft. than it does now ? Ans. 0.9454 second. 

(401) What do you understand by center of gravity ? 

(402) What do you understand by specific gravity ? 

(403) {a) What is the density of a cubic "foot of a body 
occupying a space of 800 cu. in., and weighing 500 lb.? (d) 
What is its specific gravity ? A i ('^^ 33.582. 

I(^) 17.28. 

(404) A body has been falling freely for 5 seconds; what 
is its velocity ? 

D. 0. 111.-27 



572 ELEMENTARY MECHANICS. 

(405) Suppose that a body, under certain conditions, were 
to fall freely for 3 seconds, and then fall uniformly with the 
velocity it had at the end of the third second for 6 seconds 
longer, how far would it fall ? Ans. 723.6 ft. 

(406) The weight of the head and piston of a steam ham- 
mer, together with the piston rod, is 8 tons. If it falls 8 ft. 
and compresses a mass of iron ^ in., what is the force of the 
blow ? Ans. 1,536 tons. 

(407) Explain what you understand by centrifugal force. 

(408) If a cast iron sphere 4 in. in diameter be revolved 
in a circle, in which the distance from the center of the 
sphere to the center of the circle is 15 in., what will be the 
tension of the string, the sphere making 60 revolutions per 
minute ? Ans. 13.38 lb. 

(409) The outside diameter of an engine fly-wheel is 80 
in. ; width of face, 26 in. ; average thickness of rim, 5 in. ; 
revolutions per minute, 175; what is the centrifugal force 
tending to burst the rim ? Ans. 38,641 lb. 

(410) If a body weighs one pound at a distance of 100 
miles from the center of the earth, (a) what will it weigh at 
the surface ? (d) At 100 miles above the surface ? Take the 
earth's radius as 4,000 miles. * j {a) 40 lb. 

^^' 1 (d) 38.0721b. 

(411) What would be the horsepower of a machine that 
could raise 10,746 lb. 354 ft. in 10 minutes ? 

Ans. 11.5275 H. R 

(412) How far above the surface of the earth will a 2-lb. 
ball weigh 3 oz. ? Ans. 9,064 miles. 

(413) What is the range of a projectile thrown horizon- 
tally, 50 ft. above a level plain, the initial velocity being 140 
ft. per second ? Ans. 246.87 ft. 

(414) A projectile has an initial velocity of 30 ft. per 
second. How far below the horizontal line of direction will 
it strike a body 10 ft. away ? Ans. 1 ft. 9.44 in. 

(415) What do you understand by moment of a force ? 
Illustrate it. 



ELEMENTARY MECHANICS. 57:3 

(410) Illustrate your idea of a couple. 

(417) Can a couple have a single resultant force ? 

(418) Where is the center of gravity of a triangle located ? 

(419) li A B C D\^ 3. quadrilateral, and A B — 10 in., 
B C=S in. , C Bf=7 in. , D A = 9 in. , and the angle be- 
tween A B and B C is 90°, where is the center of gravity of 
the figure ? Determine it graphically. 

(420) AVhere is the center of gravity of a regular penta- 
gon, the length of a side being seven inches ? 

(421) If the weight of the balls shown in Fig. 128 were 
W, = 21 lb., W^ = 15 lb., PF3 = 17 lb., and W^ = 9 lb., where 
would the center of gravity be, the distance between the 
centers of IV^ and W^ being 34 in. between IV^ and W^, 25 
in. ; between W^ and IV^, 40 in. , and between IV^ and IV^, 
18 in.? 

(422) Find the center of gravity of a square board of 
uniform thickness whose sides are 14 in. in length, and having 
one of its corners cutoff at a distance of 4 in. , measured 
from that corner each way ; or, what is the same thing, 
cutting off a right-angled triangle whose sides, including the 
right angle, are 4 in. long. 

(423) A bookbinder has a press, the screw of which has 
4 threads to the inch. It is worked by a lever 15 in. long, to 
which is applied a force of 25 lb. ; (a) what will be the pres- 
sure if the loss by friction is 5,000 lb.? (d) What would be 
the theoretical pressure ? (c) What is the efficiency in this 
case? f (a) 4,424.8 lb. 

Ans. ] (d) 9,424.8 lb. 
( (c) 46.95^. 

(424) How would you determine whether a body was 
stable or not if placed in a certain position ? 

(425) If a prism 10 inches square has been so cut that its 
axis is 22 inches long and makes an angle of 60° with the 
base, will the prism stand or fall when placed on its base ? 
Consider the plane of the upper base as being at right 
angles to the axis. 



574 



ELEMENTARY MECHANICS. 



(42G) If the force moves through a distance of 5 ft. 6 in., 
while the weight is moving 6 in., {a) what is the velocity- 
ratio of the machine ? (d) What weight would a force of 
5 lb. applied to the power arm raise ? 

(427) In the last example, if the efficiency were 65fo, what 
weight could be raised ? 

(428) The length of a lever is 5 ft. ; where must the ful- 
crum be placed, so that a weight of 35 lb. at one end may 
balance one of 180 lb. at the other end ? 

(429) In a block and tackle the theoretical force neces- 
sary to raise a weight of 1,000 lb. is 50 lb. ; {a) what is the 
velocity ratio ? (d) How many pulleys are there ? (c) If the 
actual force necessary to raise the load is 95 lb., what is the 
efficiency? Ans. (^) 52.63^. 

(430) The nuts on a cylinder head are tightened by a 
wrench whose handle is 20 in. long. If the force exerted 
upon the wrench is 60 lb., and the efficiency of the combina- 

-tion is 40^, what pressure will the nut exert against the head 
(or, in other words, what is the tension of the sttid), the 
pitch of the screw being ^ of an inch ? Ans. 24,127.5 lb. 

(431) The base of an inclined plane is 20 ft. in length and 
its height is 5 ft. ; (a) what force acting parallel to the plane 
will balance a weight of 1,580 lb.? (d) What force acting 
parallel to the base would balance this weight ? 



Ans. 



J(^) 383.2 lb. 
( (d) 395 lb. 



(432) Find what the 
weights W and W must be 
to produce equilibrium 
when the levers shown in 
Fig. 1 are suspended from 
the ring A. 

(433) If in Fig. 141, the 
power arms P F equal 14, 
21 and 19 inches, respec- 
tively, and the weight arms 



12" 



WL 




U2lb, 



Fig. 1. 



ELEMENTARY MECHANICS. S'i'S 

W F equal 2.V, 3:^ and 2|- inches, respectively, what force 
applied at P will raise a load of 725 lb. ? Ans. 3.032 lb. 

(434) In Fig. 144, suppose the radius of the wheel A is 15 
in. ; of C, 12 in., of £, 20 in. ; of the drum F, 5 in. ; of the 
pinion D, 3^ in., and of B, 3 in. ; {a) what load would a force 
of 35 lb. applied at Praise ? (l?) What is the velocity ratio ? 
(c) If the weight actually raised was 1,932 lb., what is the 
efficiency? Ans. Efficiency =: 80.5^. 

(435) A frame having the shape of an equilateral triangle 
measuring 15 in. on each edge is suspended in a horizontal 
position, weights of 12, 15, and 18 lb. , respectively, being hung 
from each corner. Where is the point of suspension that the 
frame may remain horizontal ? Solve graphically, and 
measure the perpendicular distances from the point of suspen- 
sion to each edge of the triangle. 

(436) A stone weighing 500 lb. is balanced on the edge of 
the roof of a building 75 ft. high; {a) what is its potential 
energy ? (/;) If all of its potential energy could be changed 
into kinetic energy, without any loss through friction or heat, 
how many horsepower would be developed, on the supposition 
that the work was done in the same time that it would take 
the stone to fall freely to the ground ? Ans. (d) 31.57 H.P. 

(437) A cubic foot of a certain kind of stone weighs 127 
lb. ; what is its specific gravity ? Ans. 2.032. 

(438) The specific gravity of bismuth is 9.823; what is 
the weight of a cubic inch ? Ans. .3553 lb. 

(439) In the differential pulley shown in Fig. 149, the 
radius of the larger pulley is 6|- in., and of the smaller pulley, 
5f in. ; what weight will a force Pof 60 lb. raise if the effi- 
ciency of the mechanism is 48€ ? Ans. 499.2 1b. 

(440) If a hammer whose head weighs l^lb., strikes a 
nail with a velocity of 25 ft. per second, driving it f of an 
inch into the wood, what is the force of the blow ? 

Ans. 466.42 lb 

(441) If 4 cu. ft. of copper alloy weigh a ton (2,000 lb.), 
{a) what is its specific gravity ? {I?) What is the weight of a 
cu. in. ? Ans. (a) Sp. Gr. 8. 



576 ELEMENTARY MECHANICS. 

(442) If the distance between the center hne of the handle 
and the axis of the drum shown in Fig. 143 is 14^ in., and 
the diameter of the drum is 5 in., what load will a force of 
30 lb. exerted on the handle at P raise ? 

(443) If the coefficient of friction is .21, what force would 
be required to move a body weighing 75 lb. ? 

(444) If a man raises a weight of 900 lb. 150 feet in 15 
minutes, by ineans of a fixed and movable pulley, {a) how 
much work has he done ? {U) What part of a horsepower is 
this equivalent to ? Ans. (U) y^ H.P. 

(445) In the last example, what horsepower would the 
man have actually expended if the resistance due to friction 
had been 36^ of the load ? Ans. .3709 H.P. 

(446) If a force of 18 lb. is just sufficient to move a weight 
of 88 lb. along a horizontal plane, what is the coefficient of 
friction ? Ans. .2045. 

(447) If 3 cu. ft. of a certain material weigh 1,200 lb., 
what is its density ? Ans. 12.438. 

(448) An iron plate rests upon four supports; upon it is 
placed a weight of 125 lb. ; a compressed spring, placed under 
the plate directly under the center of gravity of the plate 
and weight, exerts an upward pressure of 47-J- lb. What is 
the pressure upon each support ? Neglect weight of plate. 

(449) Find the resultant of the 
forces acting in Fig. 2 — all acting 
towards the same point. 

(450) The distance between 
the center line of the handle and 
the axis of the drum in Fig. 143 
is 12 in., and the diameter of the 
drum is 4^ in. The free end of 
the rope, forming part of a block fig. 2. 
and tackle having 6 pulleys, is 

wound up on this drum. How great a weight can be 
lifted by the pulleys if a force of 30 lb. is exerted on the 
handle ? Ans. 960 lb. 




ELEMENTARY MECHANICS. 577 

(451) (a) What is the velocity ratio in the last example ? 
(d) If the weight actually lifted by a force of 30 lb. was 
790 lb., what is the efficiency ? a i (^) 32. 

( (l?) 82.29^. 

(452) It is desired to raise a weight by means of a pulley 
fixed overhead, the free end of the rope passing over another 
pulley, fixed to the floor. If the resistance due to friction is 
24^ of the load lifted, (a) what force would be necessary to 
raise a weight of 475 lb. ? {d) What is the efficiency ? 

i (a) 589 lb. 
* (d) 80.64^. 

(453) Explain your idea of work, power, horsepower, and 
kinetic energy. If a constant force of 6 pounds can cause a 
body weighing 60 pounds to move a distance of 25 feet in 2^ 
seconds, what is (a) the work done ? (<^) The power expended ? 



STRENGTH OF MATERIALS 

(ARTS. 1331-1421.) 



EXAMINATIOiV QUESTIONS. 

(689) (a) What are the advantages and disadvantages of 
cast iron as a material for construction purposes ? {I?) Of 
steel ? {c) Of wrought iron ? (d) For what purposes are each 
of the above largely used ? 

(690) Define the following: (a) Stress; (^) strain; (c) 
elasticity; (d) coefficient of elasticity; (e) ultimate strength. 

(691) In how many ways may stress be applied to a body ? 
Give an example of each. 

(692) How much will a wrought-iron rod ^" in diameter 
and 10 feet long be elongated by a pull of 40 tons ? 

Ans. .12223 in. 

(693) A steel rod 7Y loi^g ^^^ V i^ diameter is elongated 
.009* by a pull of 7,000 pounds; what is the coefficient of 
elasticity? Ans. 29,708,853 lb. per sq. in. 

(694) What force is required to produce an elongation of 
,006^^ in a cast-iron bar lY X 2" and 9 feet long ? 

Ans. 2,500 lb. 

(095) A wooden rod S" in diameter is elongated .05" by a 
force of 2,000 pounds; what was its original length ? 

Ans. 265.07 in. 

(696) What is the minimum diameter of a wrought-iron 
bolt which is to withstand a steady pull of 6 tons with 
safety ? Ans. 1.054 in. 

(697) How long must a cast-iron bar be, if supported 
vertically at its upper end, to break under its own weight ? 

Ans. 6,400 feet. 

For notice of the copyright, see page immediately following the title page. 



1X76 



STRENGTH OF MATERIALS. 



(698) The diameter of a wrought-iron bolt is f; what 
should be the thickness of the bolt head in order that the 
bolt may be equally strong in tension and in shear ; in other 
words, what should be the thickness of the head in order 
that the force required to pull the bolt apart shall be just 
equal to the force required to strip the head from the bolt ? 

Ans. .206 in. 

(699) What safe load may be borne by a brick pier, the 
cross-section of which is 2|- X 3^ feet ? Ans. 105 tons. 

(700) In Fig. 11, the force P is If tons and acts at an angle 

of 30° with the horizon- 
tal. The width b of 
the timber tie-rod is 4 
inches. Find the safe 
length a necessary to 
prevent shearing on the 
surface a b. 

Ans. 10.1 in. 

(701) Show that a 
cylindrical boiler is 
twice as liable to rup- 
ture along a longitudinal seam as along a transverse seam. 




Fig. 11. 



(702) Assuming that the strength of the boiler plate is 
reduced 40^ owing to the area lost by the rivet holes, what 
should be the thickness of the plate of a wrought-iron boiler 
4 feet in diameter which sustains a steam pressure of 120 lb. 
per sq. in. ? Assume factor of safety for steady stress. 

Ans. .35 in. 

(703) What should be the thickness of a 6" cast-iron 
water pipe to carry a steady pressure of 200 lb. per sq. in. ? 

Ans. .18 in. 

(704) What should be the thickness of a wrought-iron 
boiler tube 3" in diameter, 12 ft. long, and exposed to an 
external steam pressure of 130 lb. per sq. in. ? Use a factor 
of safety of 10. Ans. .272 in. 



STRENGTH OF MATERIALS. 



1177 



(705) A cast-iron cannon of 4" bore is subjected to an 
internal pressure of 2,000 lb. per sq. in. on being fired; what 
should be the thickness of the metal in order that it shall not 
be subjected to a stress of more than 2,800 lb. per sq. in.? 

Ans. 5 in. 

(706) A structure of the nature of a simple beam is 100 
feet long between the supports, and carries three equal 
loads of 1,200 lb. each, at distances from the left support of 
40, 60, and 80 ft. (a) Find the maximum bending moment; 
(l?) the shear at a distance of 30 ft. from the left support ; 
(c) the maximum shear. ( (a) 748,800 in. -lb. 

Ans. ] {d) 1,440 lb. 

( (c) -2,160 1b. 

(707) Find graphically the magnitude, direction, and 

position of the resultant of the forces shown in Fig. 12. 

Note. — In Figs. 12 and 14, the line 7nn is drawn simply to locate 
the forces. Thus, in Fig. 12, i^i makes an angle of 47|° with m n ; Fi 







"^^^: 

^^' F. 



■n 



Fig. 12. 



makes an angle of 72-^° with m n, and so on. The distance between 
tiie intersection of Fi and F<i with ;;z ;/ is 1 inch ; the distance between the 
intersections of F2 and F2 is 1^ inches, etc. 

(708) A beam is loaded 
as in Fig. 13. Find the 
reaction of the supports, 
and draw the shear diagram. 

(709) {a) Where is the 
maximum bending moment 
of the above beam ? {b) 
What is it in inch-tons ? 
(c) In inch-pounds ? 

\{b) 288 in. -tons. 




Ans. 



Fig. 13. 



( {c) 576,000 in. -lb. 



1178 



STRENGTH OF MATERIALS. 



(710) Define the following: Simple beam: cantilever; 
restrained beam ; continuous beam. 




Xfir-- 



Fig. 14. 

(711) Find the resultant moment of the forces of Fig. 
14 about the point C. 

(712) A rectangular white-pine beam 22 feet long car- 
ries a load of 4 tons, 8 feet from one end. Assuming 
the depth of the beam to be double the width, what should 
be its dimensions to safely support the load ? 

Ans. Depth =: 17|- in. ; breadth = 8f in. 

(713) The beam of Fig. 15 is loaded as shown and 
has, in addition, a uniform load of 40 pounds per foot. 

{a) Find the maximum 
shear; {b) the maximum 
bending moment; (<:) 
the reactions of the sup- 
ports. 

^ {a) 2,930 lb. 
\b) 320,000 in. -lb. 
\c) 2,930 and 
FIG. 15. 2,8701b. 




Ans. 



(714) The moment of inertia of a wrought-iron I beam 
12'' deep is 280 ; what uniform load will it sustain ? The 
stress is assumed to be steady. The beam is 20 ft. long and 
rests upon two supports. Ans. 875 lb. per ft. 



STRENGTH OF MATERIALS. 1179 

(715) A hollow circular cast-iron beam 8 feet long rests 
upon two supports. The inside diameter is 5 in. and the 
outside diameter 6J in. ; what is the maximum safe load 
that may be concentrated at its center ? Ans. 4,G24 lb. 

(71G) A hemlock floor beam is 2" X 10" and 16 feet long; 
(a) what distributed load will it carry ? (I?) What dis- 
tributed load would it carry if laid with the lO'' side 
horizontal? ^^^ { {a) 78.12 lb. per ft. 

^^' \{l?) 15.6 " 

(717) Find the deflections of the beams of (a) example 
711, (<^) example 715, and {c) example 716 (a). 

f (a) .45 in. 
Ans. I {^) .1 in., nearly. 
( (c) .461 in. 

(718) The piston of a steam engine is 14 in. in diameter; 
the steam pressure is 80 lb. per sq. in. Assuming that the 
total pressure on the piston comes on the crank-pin at the 
dead points, and considering the crank-pin as a cantilever 
uniformly loaded, what should be its diameter if 4 inches 
long and made of wrought iron? Take a factor of safety 
of 10. Ans. 3.82 in. 

(719) What load can be safely sustained by a cast-iron 
column 14 feet long and 6'' in diameter, with flat ends? 

Ans. 120,872 lb. 

(720) What should be the dimensions of a piece of tim- 
ber 30 ft. long, with flat ends, to support a load of 7 tons, 
the cross-section being in the form of a square ? 

Ans. 9f in. square. 

(721) A cylindrical steel connecting-rod 7^ ft. long is 
subjected to a maximum stress of 21,000 pounds. Assuming 
it to be a column with both ends hinged, and subjected to 
shocks, determine its diameter to the nearest i". Ans. 2f in. 

(722) A wrought-iron piston rod has a diameter of 2" 
and a length of 4 feet. Assuming it to be a column with 
one end flat and the other rounded, what is the allowable 
diameter of the piston if the steam pressure is 60 pounds 
persq. in.? The rod is subjected to shocks. Ans. 15J in« 



1180 STRENGTH OF MATERIALS. 

(723) A beam is 6" X 8" and 10 feet long, the 8" side 
being vertical. Another beam of the same material is 
4" X 12" and 16 feet long, the 12'' side being vertical, {a) 
What is the ratio between the maximum loads the two beams 
are capable of supporting ? (d) What is the ratio between 
the deflections in the two cases, the manner of loading 
being the same ? . ^^ j (a) lj\ : 1. 

' l(^) .549 : 1. 

(724) (a) What should be the diameter of a round 

wrought-iron shaft to transmit 40 H. P. at 120 R. P. M. ? 

{d) To transmit 80 H. P. at 100 R. P. M. ? 

^^((.) 3. 739 in. 

i(d) 4.65 m. 

(725) What must be the diameter of a steel engine shaft 
to transmit 4,000 H. P. at 50 R. P. M.? Ans. 14.22 in. 

(726) What H. P. can be transmitted by a steel shaft 4* 
square, making 80 R. P. M. ? Ans. 71.775 H. P. 

(727) What horsepower can be transmitted by a hollow 
wrought-iron shaft making 100 R. P. M., the outside 
diameter being 7^- in., and the inside diameter 5 in.? 

Ans. 717.7 H. P. 

(728) {a) What weight may be safely sustained by a hemp 
rope 8" in circumference ? (i?) What should be the diameter 
of an iron wire rope of 7 strands to safely sustain a stress of 
6,000 pounds ? (c) What should be the circumference of a 
steel hoisting rope which is required to lift a load of 6f tons? 

( (a) 6,4001b. 
Ans. ] (l?) 1.054 in. 
( (c) 3.65 in. 

(729) (a) An open link chain is made of |-" iron ; what 
safe load will it sustain ? {d) What should be the diameter 
of the iron composing the links of a stud-link chain which is 
to sustain a load of 4 tons ? a \ {^) 9,187.5 lb. 

^^' t(^) .667 in. 

(730) A steel shaft 2" in diameter is supported by hang- 
ers 10 feet apart ; if a pulley be placed midway between the 
hangers, what should be the maximum allowable belt pull on 



STRENGTH OF MATERIALS. 



1181 



the pulley in order that the shaft shall not deflect more 

thanf' ? Ans. 327.25 lb. 

Suggestion. — Assume the shaft to be a restrained beam, loaded in 
the middle. 

(731) An engine shaft rests on bearings 5 feet apart; 
midway between the bearings the shaft carries a fly-wheel 
weighing 3 tons, (a) Find the size of the shaft to withstand 
this load, {b) Find the size of the shaft on the assumption 
that the engine makes 80 R. P. M. and develops 75 H. P. 
The shaft is made of steel. Factor of safety, 10. 

( (d) ^ in., 

(732) A white oak cantilever is loaded as shown 
16, and has in addition a 
uniform load of 80 pounds 
per foot, {a) Determine 
graphically the shear d i a- 
g r a m and the maximum 
bending moment, (l?) Find 
the necessary dimensions of 
the cross-section of the beam, 
assuming the depth to be 2^ times the breadth, and the stress 
to be steady. . Ua) 82,320 in. -lb. 

^ • {d) ^ = 3.7 in 



nearly, 
nearly. 

Fig. 



m 




2^6'^ 






-1r 



F, 



CO 



^* 



^^ 



Fig. 16. 



L 8'-^ 



d= D^in. 

(733) A steel beam of the cross-sec- 
tion shown in Fig. 17 rests upon two 
supports 35 feet apart; (a) what con- 
centrated load will it sustain at the 
center, the stress being considered 
(d) What total uniform 
sustain, the stress being 
^^((^) 7,246 lb. 
I (d) 20,2901b. 

(734) {a) The moment of inertia of a rectangle is 72; the 
area of the rectangle is 24 sq. in. ; what is the value of r in 
formula 115? {b) What are the dimensions of the rect- 
angle ? (c) The moment of inertia of a circle about an axis 




as varying ? 
load will it 
steady ? 



1182 STRENGTH OF MATERIALS. 

, . . TZ d^ . . , . . . "K d^ 

tnrouQ^h its center is-—-—; the area ot the circle is — r— ; 

what is the radius of gyration of the circle ? 

[ (a) 1.732. 
Ans. \ \b) d = 4: in, ; d= 6 in. 
( (c) id 

(735) A cast-iron sphere 8" in diameter is subjected to 
a steady internal pressure of 100 lb. per sq. in. ; what should 
be its minimum thickness to safely withstand this pressure? 

Ans. .06 in. 

(736) A beam 28 feet long, uniformly loaded with 
60 pounds per foot, rests upon two supports which are placed 
5 feet from each end. Determine graphically the maximum 
shear and the maximum bending moment. 

Ans. Bending moment =20,160 in. -lb. 

(737) Assuming the above beam to be made of white pine, 
find the dimensions of its cross-section, the stress being 
steady. The beam is to be rectangular. 

(738) A wrought-iron boiler tube 2^" in diameter and 
9 feet long has a thickness of .2" ; what external pressure can 
the tube withstand, using a factor of safety of 10 ? 

Ans. 106.43 lb. per sq. in. 

(739) A crane chain is required to lift 5 tons; what 
should be the diameter of the iron composing the links of 
the chain, which is of the open link variety ? Ans. .913 in. 

(740) A gear-wheel 4 feet in diameter is keyed to a shaft. 
The force acting tangent to the pitch circle of the gear-wheel 
is 350 pounds ; what should be the diameter of the shaft if 
made of steel ? Ans. 2.84 in., say 2f in. 

(741) The head of an engine cylinder, 12* in diameter, is 
fastened on by 10 wrought-iron bolts. In order to make the 
joint steam tight, a safe stress of only 2,000 pounds per sq. in. 
is allowed. The steam pressure being 90 pounds per sq. in., 
what should be the minimum diameter of the bolts, that is, 
the diameter at the root of ihe thread ? Ans. .8 in- 



STRENGTH OF MATERIALS. 



1183 



18^ 



(742) A beam 18 feet long has a load of 1 ton placed 
at each end. The supports 
are placed 5 feet from each 
end. (See Fig. 18.) {a) 
Draw the shear diagram and 
find the maximum bending 
moment, {b) If made of 
cast iron and circular in 
cross-section, what should be its diameter, the stress being 



-$^ 



t_^^ 



Fig. 18. 



steady ? 



Ans. 



{a) 120,000 in. -lb. 
\b) 5.784 in. 



(743) A beam %" X 6" in cross-section and 18 feet long 
deflects .3"; how much will a beam 3" X 8" and 12 feet long 
deflect under the same load ? The long side is placed 
vertically in both cases. Ans. .025 in. 

(744) A wrought-iron key, or cotter, is used to fasten a 

rod, as shown in Fig. 19. There 
is a pull of 20,000 pounds on the 
rod. Assuming the breadth of 
the key to be four times the thick- 
ness, find its dimensions to safely 
resist the stress. Use a factor of 
safety of 10. 

^^g ( Breadth, 2.828". 
Thickness, .707^ 




'I 



Fig. 19. 



(745) A steel engine shaft 12" 
in diameter, resting in bearings 
54" apart, carries a fly-wheel 
weighing 30 tons, midway between 
the bearings ; what is the deflection of the shaft ? Consider 
the shaft as a simple beam. Ans. .0064 in. 

(746) {a) A 7-strand steel wire rope has a diameter of 
IJ"; what load will it carry safely ? {b) Find the circumfer- 
ence of a hemp rope to carry If tons. 

I {b) 5. 92 in. 



D. 0. III. 



28 



1184 STRENGTH OF MATERIALS. 

(747) What stress would be required to rupture a cast- 
iron cylinder which is 8" in diameter and 6" thick ? 

Ans. 12,000 lb. 

(748) A beam which is 100 feet long carries four equal 
loads of 1,200 pounds each at distances of 20, 40, 60, and 
80 feet from the left support, (a) Find the maximum bend- 
ing moment, and (d) the shear at each support. 

j (a) 804,000 in. -lb. 
' 1 (i?) 2.400 lb. 



APPLIED MECHANICS 

(ARTS. 1422-1588.) 



EXAMINATION QUESTIONS. 

(749) A lever is required to transfer motion from one 
line to another one parallel to it, and 4^ distant. The 
motion along one line is to be 3|-", and along the other line 
2f "o Determine the length of the lever arms graphically, 
assuming the fulcrum to be between the ends of the lever. 

(750) A motion of 6 inches along one line is to be trans- 
ferred and increased to a motion of 8 inches along another 
line, making an angle of 45° with the first line. The direc- 
tion of the motion is to be reversed. Make a drawing of a 
bell-crank lever, one-fourth size, which will accomplish this. 

(751) Design a bell-crank lever like that shown in Fig. 
347. Given, angle between center lines of motion, 10° ; ratio 
of motions, 4:3; length of long arm, 16 inches. Draw 
one-fourth size. 

(752) Design a lever indicator reducing motion which 
will give an exact reduction. Stroke of engine, 3 feet; 
length of card, 3^ inches. 

(753) A vibrating toothed pinion 12^ in diameter gears 
with a rack which has a corresponding reciprocating motion. 
How far must the rack move to cause the wheel to vibrate 
through an angle of 6^° ? Ans. .08 inch. 

(754) Draw a skeleton diagram of the mechanism shown 
in Fig. 20; i. e., a diagram showing the positions of the 
center line of each link when in mid-position and in each 
extreme position. Crank A is the driver ; the bell-crank B 
has equal arms of any convenient length ; the long arm of 
the lever C makes an angle of 30° with the vertical when in 

For notice of copyright, see pa^e Irnmedi^tely following the title p^ge. 



1186 



APPLIED MECHANICS. 



mid-position, and on its upper end carries a segment, gear 
ing with the pinion D. The pinion is to vibrate exactly 




Fig. 20. 



one-half a turn for each stroke of the lever C. All necessary 
dimensions are given in the figure. 

(755) In the connecting-rod mechanism of Fig. 349, (a) 
does the cross-head move more rapidly during the first or 
during the second half of the return stroke ? [b) What 
effect does increasing the length of the connecting-rod have 
upon this motion ? 

(756) Two cranks of the same length are keyed to the 
same shaft so as to be parallel, and revolve together, side 
by side ; one crank drives a cross-head A by means of a 
connecting-rod, and the other crank drives a slotted cross- 
head B. If the crank turns left-handed and cross-head A is 
at the left of the crank-shaft, {a) which cross-head leads 
during the forward stroke ? (U) Which, during the return 
stroke ? 



APPLIED MECHANICS. 



1187 



(757) In Fig. 21 is a bookbinder's press, consisting of two 
equal armed toggle-joints, drawn together by a right and 




Fig. 21. 

left hand screw. U R= 10", L = 18", A = 12\ and the 
screw has four threads per inch, what pressure would the 
platen P exert when a force of 75 lb. is exerted at the end 
of the handle, and the distance B is 16 inches ? Neglect 
friction. Ans. 84,295.4 lb. 

(758) Fig. 22 illustrates a mechanism in a certain 
machine. A pulley P drives shaft O^ to w^hich is attached 
crank disk /^, the radius of the crank being OR. C P is a. 
vibrating lever with a pin P, at the end of which the con- 
necting-rods R P and K P are attached, thus transmitting 
the motion to the cross-head H. With a belt pull of 2 lb., 
it is desired to find (a) the force exerted upon the pin K, 
and (d) the horizontal thrust exerted by the cross-head //, 
when C P makes an angle of 30° with the horizontal. 

As the example is to illustrate the principle only, the 
dimensions are, for convenience, made small. 



11! 



APPLIED MECHANICS. 



Given, diameter of pulley, 3"; radius O R = ^ \ radius 
C E^ If" ',KE= 5V'. The length of i^ ^ is such that the 



crank disk will cause point E to move exactly to E' . 




Fig. 22, 

Note. — Draw OE and consider EO R as a crank and connecting- 
rod mechanism, with the cross-head at E^ and moving in the direction 
of the arrow. 

Ans \^''^ ^^•^^^^• 
( {b) 13.86 lb. 

(759) What adjustments to the stroke are necessary 
in a shaping or slotting machine where the tool has a 
reciprocating motion ? 

(760) {a) Show graphically how to construct a vibrating 
link motion, so that the periods of advance and return are 
to each other as 3 is to 2. Length of stroke, 24"; scale, 
3" = 1 ft. (U) What would be the effect of shortening the 
stroke of the tool slide ? 

(761) {a) Show how to construct a Whitworth quick- 
return motion, so that the period of return equals one-half 
of the period of advance. [U) Plot the motion for positions 
of the driving gear at every 20° of its motion, {c) What do 



APPLIED MECHANICS. 1189 

you consider to be the relative merits of the vibrating link 
and Whitworth quick-return motions ? 

(7G2) (a) A universal joltit is used to couple two shafts at 
an angle of 28° with each other. If the driving shaft re- 
volves 50 times per minute, what would be the greatest and 
least rates of motion of the driven shaft ? (d) What would 
be the effect of using two universal joints, if the forks on 
the middle shaft which connects them were at right angles 
with each other ? 

Ans. {a) Greatest = 5(3. G3 rev. Least = 44 15 rev. 

(7G3) Given the stroke of a Watts straight-line motion 
= 2 ft. ; distance from the center of the lower lever to the 
center line of motion of the guided point = 30"; perpendic- 
ular distance between the two levers when in mid-position 
= 3 ft. ; perpendicular distance of the guided point from 
the lower end of the connecting link = 15''. Required the 
lengths of the two levers. 

. j Length of upper lever, 23.1". 
I Length of lower lever, 31.2". 

(704) Find the outline of a cam to turn uniformly right 
handed and give motion to a point moving on a line pass- 
ing through the axis of the cam. The point is to remain 
stationary during the first -^^ revolution of the cam ; to move 
uniformly away from the center of the cam 2" for the next 
y\ revolution, and then to move to its first position with a 
harmonic motion during the remaining y^g- revolution. Take 
\y as the shortest distance between the point and the axis 
of the cam. 

(765) Draw the outline for a cam to turn uniformly left- 
handed, and give a motion to a roller \" in diameter in a 
horizontal line, passing V above the axis of the cam. The 
roller is to have a uniform motion of 2^' to the left during 
the first \ revolution of the cam; during the next \ revolu- 
tion it is to remain at rest, and then is to move suddenly to 
its first position, where it is to remain during the last quar- 
ter-turn of the cam. The nearest position of the center of 
roller to the axis of the cam is to be %" . 



1190 APPLIED MECHANICS. 

(7GG) A cylindrical cam d" in diameter is designed to 
move a roller uniformly to the left a distance of 1" during 
the first half revolution ; to move it uniformly to the right Y 
during the next ^ revolution, at which point it remains dur- 
ing the next quarter revolution. During the last -J- of a 
revolution, it returns uniformly to its first position. The 
cam turns left-handed as looked at from the right. Draw 
the development of the groove, the diameter of the roller 
being Y. 

(767) Lay out the outline for a positive-motion plate cam, 
like that in Fig. 369. Diameter of rollers, f"; distance 
between centers of rollers, 5" ; required movement of rod, 1:J^". 

(768) A pulley 4 ft. in diameter drives another pulley 52" 
in diameter by means of a belt. If the speed of the driven 
shaft is 320 revolutions per minute, what is the speed of the 
driving shaft ? Ans. 346|- rev. per min. 

(769) The driving pulley on a machine is 10" in diameter, 
and should run at 500 revolutions per minute. What should 
be the diameter of the pulley on the countershaft, the speed 
of which is 210 revolutions per minute ? Ans. 23|-". 

(770) A saw arbor is fitted with a pulley 3|- inches in 
diameter. The countershaft has pulleys 20" and 6" in diam- 
eter. What size pulley should be placed on the main shaft, 
which runs at 189 revolutions per minute, to drive the saw 
at 1,800 revolutions per minute ? Ans. 10 inches. 

(771) A machine is to be belted through a countershaft 
so as to run at 900 revolutions per minute, the speed of the 
main driving shaft being 150 revolutions per minute. Find 
two ratios that could be used for each pair of pulleys. 

(772) A shaft making 200 revolutions per minute has 
upon it a pulley 36" in diameter, carrying a belt upon its 
periphery which transmits 5 horsepower, (a) What is the 
effective pull on the belt ? (d) What diameter should the 
pulley have in order that the effective pull may be 50 lb. ? 

Ans. ] W «7.5-nb. 

( (d) 63 inches +. 



APPLIED MECHANICS. IIDI 

(773) Determine by the aid of Table 33 what you would 
consider to be a suitable width of single belt for driving a 
saw arbor under the following conditions : Greatest horse- 
power required, G; diameter of smaller pulley, 4"; diameter 
of driving pulley, 30" ; distance between the centers of the 
two pulleys, 5 ft. ; number of revolutions ot the saw arbor, 
1,500. 

(774) A 2o0-horsepower Corliss engine makes 90 revolu- 
tions per minute, and has a 14-ft. band wheel. How wide 
should the rim of the wheel be, the rim being 1" wider than 
the belt. Calculate for a double belt. Ans. 41 inches. 

(775) Two shafts connected by a G-inch single belt run- 
ning over 24-inch pulleys make 205 revolutions per minute. 
How much power should the belt transmit ? 

Ans. 8.G H. P., nearly. 

(776) If the belt were found to be inadequate for the 
work in the last example, which would be the more effectual 
remedy — to use a G-inch double belt in place of the single 
belt, or to substitute 30" pulleys for the 24" pulleys, and still 
use a 6-inch single belt ? 

(777) Two continuous cones are required to give a range 
of speed between 70 and 400 revolutions per minute. As- 
suming the large diameters of the cones to be 18", (a) what 
must be the small diameters ? (d) What should be the speed 
of the driving shaft ? Both cones are to be alike. 

^^((.) 7.53 in. 
I {d) lG7i rev, 

(778) In the above example, if the speed of the driving 
shaft were 225 revolutions per minute, (a) what should be 
the ratio of the large and small diameters, assuming the 
highest speed of the driven cone to be 400 revolutions per 
minute ? (d) Thus designed, what would be the slowest 
speed at which the driven cone could run ? 

Ans. I W 1-^8:1. 

( (d) 12G.5Grev. 

(779) A pair of stepped cones, having 4 steps, each of 
2-inch face, are required to be used with an open belt. 



1192 APPLIED MECHANICS. 

Given, largest diameter, 15"; smallest diameter, 4:^ ; dis- 
tance between centers, 20"; width of each step, 2". Make 
an outline drawing of one of the cones, half size. 

(780) Make a sketch showing how you would lace an 
8-inch belt. Show the location of the holes by dimension 
figures. 

(781) Two shafts lie in parallel planes, one above the 
other, but cross each other at an angle of 45°. Make a 
sketch showing how they might be connected by belt. Show 
two views, either a plan and elevation, or front and side 
elevation, and indicate by arrows the directions in which the 
shafts are to run. 

(782) Two shafts lie in the same vertical plane, but make 
an angle of 30° with each other. Show how to connect 
them by belt so that they will revolve in opposite directions, 
and will run either backwards or forwards. Indicate the 
directions of rotation by arrows. 

(783) A pulley A, upon a main shaft, drives a pulley B 
by means of a crossed belt. A spur gear (7, on the shaft 
with B, drives a pinion i7. A pulley E, on the shaft with Z>, 
drives a pulley F by an open belt. 

Given, A, 30" in diameter and making 60 revolutions per 
minute; B, 15" diameter; C, 60 teeth; B, 15 teeth ; B, 30" 
diameter, and B, 10" diameter. How many revolutions 
does pulley B make per minute, and in what direction does 
it turn relatively to A ? Ans. 1,440 revolutions. 

(784) A train of four gears is arranged as follows : 
Gear A drives gear B; gear B drives gear C; gear C drives 
D. A has 90 teeth and turns right-handed; B has 40 teeth; 
C has 80 teeth, and B>, 90 teeth. How many turns does gear 
D make to every turn of A, and in what direction ? 

(785) Suppose the lead screw of a lathe to have a left- 
handed thread, the pitch being -J-", and suppose, the stud T 
(Fig. 398) to make ^ as many turns as the spindle d^^ having 
27 teeth, and/"^, 54 teeth. Find a set of change gears, the 
smallest gear to have 24 teeth, and the largest 96 teeth, to 



APPLIED MECHANICS. 



1193 



cut threads from 4 to IG per inch (including llj per inch), 
making the number of gears as small as possible. 

Threads _ ^ . , Gear on 



Ans. 



J- Xi.X V^dVAO 

per Inch. 


Gear on Stud. 


Lead Screw. 


4 


9G 


24 


5 


96 


30 


6 


96 


36 


7 


96 


42 


8 


96 


48 • 


9 


96 


54 


10 


48 


30 


11 


48 


33 


m 


96 


69 


12 


48 


36 


13 


48 


39 


14 


48 


42 


15 


48 


45 


16 


48 


48 



(786) A mangle rack and wheel are so proportioned that 
5 turns of the wheel will produce one complete forward and 
back movement of the rack. If the gear turns uniformly, 
and has a pitch diameter of 10", (a) how far will the rack 
travel one way with a iiniform velocity? (^) What is the 
total travel of the rack one way? « j {a) 5 ft. 2.832 in. 

\\b) 6 ft. .832 in. 

(787) In the clutch gearing, shown in Fig. 403, suppose 
the worm gear W to have 40 teeth; gear D^^ 80 teeth; gear 
F^, 20 teeth, and the two bevel gears to be of the same size. 
If the worm is single-threaded, what is the ratio of the 
"quick return ' through the spur and bevel gears to the slow 
motion through the worm and worm-wheel? Ans. 160 : 1. 

(788) In Fig. 404, let the diameter of D^ be 3^' ; of /^„ 26" ; 
of D^^ 3^". What is the ratio of the belt to cutting speed, sup- 
posing the driving pulleys to be 30" in diameter? Ans. 63. 6 : 1. 

(789) In the differential train shown in Fig. 406, let 
wheel D have 100 teeth and remain stationary, {a) If 
wheel F has 99 teeth, and A revolves + 10 times, how many 



1194 



APPLIED MECHANICS. 



turns will /^make, and in what direction? (d) If P should 
have 101 teeth, under the same conditions, how many turns 
would i^make, and in what direction? 



Ans 



j {a) —.101 turn. 
* ( (^) +.099 turn. 



(790) In the same train, let D have 50 teeth and F 
40 teeth, (a) If D makes -|- 1 turn and A -\- 4: turns, how many- 
turns does i^ make, and in what direction? (l?) If £> makes 
— 1 turn and A -\- 5 turns, how many turns does F make? 

Note. — Consider the train locked as usual. Fix arm and turn D 
back to the starting point; then, give D the stated number of turns, 
with the arm still fixed. c /^\ i x turn 

^^^' ] (^) _ li turns. 

(791) In Fig. 407, let F^ have 20 teeth; D^, 40 teeth; F, 
20 teeth, and Dy 200 teeth. If ^ makes + 6 turns, how 
many turns does F^ make? Ans. ~\- 126 turns. 

(792) In the differential train shown in Fig. 23, pulleys 
A and B are connected by open belt. A and F are fast to 



100 




'20 tJRM. 



shaft S\ D and H^ and /fare loose on 5; // and K bemg 
one piece. Wheel D serves as an arm to carry gears F and G. 
B and C are keyed to shaft L. 



APPLIED MECHANICS. 1195 

Let //", /% E^ and G each have. 50 teeth; D^ 100 teeth, and 
r, 10 teeth, and let the diameter of A be 12" and of B 3^ 
If shaft wS makes -j- 20 turns, as indicated by the arrow, how 
many turns will pulley iTmake, and in what direction? 

Note. — Calculate as though E were stationary at first. Then, fix 
arm D, and turn ^+20 times. ^^s. - 36 revolutions. 

(793) In designing gear teeth, what object should be 
accomplished, (a) with regard to the motion transmitted, 
and {li) with regard to the shape and contact of the teeth? 

(794) (a) What are the relative advantages of circular 
and diametral pitch? {U) The diametral pitch of a gear is 
^\\ what is the circular pitch? (<:) The circular pitch of a 
gear is 1.1424"; what is the diametral pitch? 

A_„ J (P) .698 inch. 
^"'- I {c) 2i. 

(795) (a) Of what diameter is a 3-pitch gear having 
30 teeth? (b) What should be the outside diameter of this 
gear? {c) What should be the working depth of the teeth? 

(a) 10 inches. 
\b) 10.667 inches. 
\c) .667 inch. 
(d) .709 inch. 

(796) Define addendum; face; root; flank. 

(797) (a) How many teeth should a 10-inch 48-pitch 
spur gear have ? (U) A gear blank measures 14" in diam- 
eter, and is to be cut 6 pitch. How many teeth should the 
gear cutter be set to space ? a i C'^) ^^^ teeth. 

( (b) 82 teeth. 

(798) What is the diameter of a spur gear which has 70 
teeth, and whose circular pitch is \\" ? Ans. 27.852 inches. 

(799) Given, the distance between the centers of two 
gears = 15^'. What must be their diameters, so that the 
ratio of their speeds will be as 3 is to 2 ? 

Ans. 18| and 12| inches. 

(800) (a) How long is the shortest arc of action that can be 
allowed with a pair of gears in running contact ? (b) In the 



{d) The total depth? 

Ans. 



1196 APPLIED MECHANICS. 

epicycloidal system, why is the generating circle never made 
larger than -| of the diameter of the smallest wheel of the set ? 

(801) What can you say for and against both the epicy- 
cloidal and involute systems of gearing ? 

(802) It is desired to lay out the tooth curves for an 
epicycloidal internal gear and pinion; diametral pitch, 3; 
number of teeth in gear, 30 ; number in the pinion, 21. What 
diameter-generating circle should be used ? Ans. H inches. 

(803) Determine whether two 4-pitch involute gears of 
12 and 16 teeth, respectively, would interfere when running 
together, without rounding the points of the teeth. Take 
the angle of obliquity = 15°. 

(804) Lay out two bevel-gear blanks from the following 
data: Shafts at right angles; ratio of speed, 4:3; diametral 
pitch, 4; number of teeth in largest gear, 24; length of face 
of gears, 1^. Write all necessary dimensions on the draw- 
ing. It is not necessary to draw the hubs. 

(805) A triple-threaded worm drives a worm gear having 
40 teeth. How many revolutions of the wheel will be caused 
by 100 revolutions of the worm ? Ans. 7. 5 revolutions. 

(806) Given, the distance between the centers of the 
shafts of a worm and worm-wheel = 6" ; ratio of speeds 
= 40 : 1 ; pitch of (single-threaded) worm thread = f " ; the 
thread and teeth are made according to the involute system. 
The length of the worm is to be 4:'\ and the diameter of the 
worm shaft lY- The teeth of ' the gear are to be propor- 
tioned according to Table 34, column 2. 

(a) Calculate the number of teeth, pitch diameter, out- 
side diameter, and whole depth of gear. 

(l?) Make a drawing showing the longitudinal section of 
the worm ; write all needed dimensions on the drawing. 

(807) A worm 2 inches in diameter, having 3 threads 
per inch, is to drive a spur gear having slanting teeth. 
What angle should the teeth make ? Ans. 86° 58', nearly. 

(808) A ratchet having 72 teeth is attached to the end of a 
screw having 11|- threads per inch. How much "feed " would 
the screw give for every ratchet tooth moved ^ Ans. -^^ inch^ 



MACHINE DESIGN. 

(ARTS. 1901-2003.) 



EXAMINATION QUESTIONS. 

(959) What procedure is adopted in estimating the cost 
of making a machine? 

(960) What is a shrink rule, and why is it used? 

(961) (a) What are chilled castings ? (d) What is the 
chief difficulty encountered in using cast iron ? 

(962) (a) Name the different kinds of screw threads, (d) 
Describe the United States standard (U. S. S.) thread. 

(963) Calculate the diameter at the bottom of the thread 
for the following U. S. S. screws: f', i\ IJ'', and 2" 
diameter. 

(964) (a) Calculate the number of threads per inch for 
the following U. S. S. screws: ^", f", 1", and lY diameter. 
(d) What is the pitch in each case ? 

(965) (a) What is the object of using multiple-threaded 
screws ? (3) If a triple-threaded screw has a pitch of J", 
what is the divided pitch ? 

(966) Why are triangular threads more suitable for 
fastenings than square threads ? 

(967) Calculate the nominal diameter of a wrought-iron 
bolt subjected to a varying tensile stress of 11,800 lb. 

Suggestion. — See Table 28, for proper factor of safety. 

Ans. li". 

(968) Calculate the diameter of the bolt in the last ex- 
ample by taking the value of Sf as 6,000 lb., as recommended 
in Art. 1935. Can you suggest any reason why the 
diameter here obtained should be used instead of that 
obtained in example 967 ? 

For notice of copyright, see page immediately following the title page- 



1722 MACHINE DESIGN. 

(969) Find (^) the outside diameter ; (d) the diameter at 
the root, and (c) the number of threads in the nut for a 
single square-threaded screw to transmit motion to a load of 
12,000 lb. ( (a) 2V'. 

Ans. j (d) r. 

i (c) 10 threads. 

(970) Calculate the various dimensions of a finished bolt 
and hexagonal nut, with washer, the bolt being y in diam- 
eter and 4" long under the head. Make a full-sized drawing 
snowing nut on bolt and the bolt threaded for If" of its 
length. Show a hexagonal bolt head. 

(971) Make a drawing of a wrench similar to that shown 
at Af Fig. 606, which will fit the nut in example 970. Take 
length of wrench as 16". Draw to a half size scale. 

(972) A wrought-iron eye bolt, similar to that shown in 
Fig. 612, is subjected to a varying tensile stress of 2,500 lb. 
Calculate the values of a^ b^ d^ and d^. (See Table 28, for 
factor of safety.) 

(973) (a) What is the object of using a jam-nut ? {b) 
For what reason should the smaller nut be placed at the 
bottom ? 

(974) Make a drawing of the locking device shown in 
Fig. 623 to a full size scale, taking the diameter of the bolt 
as V. 

(975) What keeps the nut from turning in the devices 
shown in Figs. 628 and 629 ? Is not the principle the same 
in both ? If so, why } 

(976) Make a drawing of the knuckle joint shown in Fig. 
642. Take d equal to 1^', and make drawing half size. 

(977) Calculate the dimensions of an ordinary sunk key 



1 \ii 



for a shaft 4" in diameter. Ans. \" X \ 

(978) Calculate the dimensions of a key for a 12" shaft 
by formula 233. Ans. 3" X 2". 

(979) Calculate the dimensions of a key for a driving 
pulley on a %-^' shaft- Ans. fj-" X i^". 



MACHINE DESIGN. 1723 

(980) Calculate the dimensions of a feather key for a 
shaft lY in diameter. Ans. ^ x f ". 

(981) A pulley transmitting 1.85 H. P. is keyed to a 
2^" shaft running at 110 revolutions per minute. Calcu- 
late the dimensions of a key for, this pulley. Ans. yV X i". 

(982) Make a drawing of a rod, cotter, and socket (all of 
wrought iron) similar to that shown in Fig. 658, assuming 
that Sf = 6,000 lb. per sq. in., and the load or pull on the 
rod = 7,000 lb. Make drawing half size, and take taper of 
cotter as -^j. 

(983) Find the diameter and length of a steel end journal 
supporting a load of 5 tons. Take S^ as 14,000 lb. per sq. 
in. and the pressure per square inch of projected area as 
400 lb. Ans. 3" X 8f ". 

(984) What should be the diameter and length of a neck 
journal for a steel shaft which is subjected to a total load of 
15,000 lb.? Take/ as 1,200 lb. per sq. in. of projected area 
and Sj> as 14,000 lb. per sq. in. Ans. 2" X 6i". 

(985) The diameter and length of a certain journal were 
calculated to be 1^" and 2^", respectively. For certain rea- 
sons it was desired to increase the length to 2Y ; what should 
be the new diameter ? Ans. IJf". 

(986) An 8" shaft is subjected to a thrust of 15,000 lb. 
Assuming 6 collars to be used, find the diameter and thick- 
ness of each. Ans. 10|-" and lyV'- 

(987) Find the diameter of a cast-iron pivot journal 
which supports a load of 1,200 lb., and turns at the rate of 
90 R. P. M. Ans. 2^". 

(988) (a) Find the diameter to which a 10" solid steel 
shaft should be increased in order that it may have a hole 
through it and still be of the same strength, assuming the 
hole to be four-sevenths of tlie outside diameter, (d) What 
is the diameter of the hole ? (c) What is the difference vx 
weights per foot in length of the two shafts ? 

( (a) lOf", nearly. 
Ans. •! (I?) 6", nearly. 
( (c) 76 lb. 

D. 0. iii.~,:<j 



1724 MACHINE DESIGN. 

(989) {a) Determine the pitch of a screw with triangulai 
threads and a diameter of 3^" ; {d) of a screw 7" in diameter, 
with square threads; (c) of a screw 12" in diameter, with 
trapezoidal threads. r {a) .3". 

Ans. ] (d) 1.4". 
( {c) 1.6". 

(990) {a) In each of the cases of the last example, deter- 
mine the number of threads per inch, (d) Calculate, also, 
the diameter of bolt at roots of thread. r (a) 2|-. 

Ans. {b) \ (6) of. 

(991) (a) Explain fully by diagram or otherwise the 
relation between the angle of the thread, the friction between 
thread and nut, and the force tending to burst the nut. 
(d) For what purposes are the triangular, the square, and the 
trapezoidal threads best adapted, and why ? 

(992) What should be the diameter' of a wrought-iron 
bolt subjected to a load varying from zero to 15,000 lb. ? 

Ans. 2i". 

(993) What steady load may be safely carried by two 
bolts, each 4" in diameter ? Ans. 80 tons. 

(994) (a) Calculate the outside diameter of the lead 
screw of a lathe which transmits motion against a pressure 
of 3,600 lb. (d) Calculate the necessary number of threads 
in the nut. ^^^ j (a) 1^ 



{b) 10 threads. 

(995) Design and draw, full size, a finished bolt with 
square head and hexagonal nut, the bolt being 6" long and 
subjected to a varying load of 3,200 lb. 

(996) Calculate the proportions of the locking arrange- 
ment shown in Fig. 625, assuming the diameter of the bolt 
to be 3y. Draw it half size. 

(997) {a) Calculate the breadth and thickness of a sunk 
key for a shaft ZY in diameter; {b) for a shaft b" in 
diameter ; {c) for a shaft 1^ in diameter. ( {a) |" x iV". 

Ans. \ {b) 1\" X W. 

(0 r X r. 



MACHINE DESIGN. 1725 

(998) A fly-wheel transmitting 1,000 H. P. at 50 R. P. M. 
is keyed to a shaft 10" in diameter by two steel keys, each lO'' 
long. Calculate the size of the keys. Ans. 2" X If". 

(999) A pulley keyed to a shaft 4^" in diameter drives 
another pulley keyed to a shaft 2f " in diameter. Calculate 
the widths and thicknesses of the two keys by f ormulas-234 
and 235. . j Driving pulley, l^V'' X f '. 

' Driven pulley, f " X yV- 

(1000) Calculate the dimensions of a sliding feather key 
for a shaft If" in diameter. Ans. j\" x f ". 

(1001) A pulley transmitting 6 H. P. is keyed to a shaft 
3|" in diameter, making 135 R. P. M. Find the breadth 
and thickness of the sunk key. Ans. f " x f ". 

(1002) A wrought-iron rod is fastened by means of a gib 
and cotter into a wrought-iron socket in the same manner as 
shown in Fig. 656. The rod is to sustain a steady pull of 
6,000 lb. Design and make a drawing. Assume a taper of 
-^, and take 5^ as 6,000 lb. per sq. in. 

(1003) Two straps each 3^" wide and f " thick are fastened 
to a rod 3" deep by a cotter and two gibs. Design the cot- 
ter and gibs and make a drawing of the arrangement. 
Taper to be ■^. See Fig. 661. 

(1004) A wrought-iron rod is fastened by a steel cotter. 
The diameter of the rod is 2|-". Calculate the dimensions of 
the cotter. Ans. 24" X f ". 

(1005) A wrought-iron piston rod is 3^" in diameter 
where it passes through the piston. Which method will leave 
the rod the stronger to resist tensile stress, cutting a stan- 
dard thread and using a nut, or using a cotter ? 

(1006) (a) Find the diameter and length of a cast-iron 
end journal subjected to a load of 24 tons, assuming the 
length to be 1^ times the diameter, and the safe stress 
4,250 lb. per sq. in. {d) What is the pressure per square 
inch of projected area ? a J (^) '^^ X 4J". 

( (d) 370 lb. 

(1007) Find the diameter and length of a wrought-iron 
end journal bearing a load of 5 tons, assuming the length 



1726 MACHINE DESIGN. 

equal to the diameter, and a safe stress of 8,500 lb. per sq. 
in. ' Ans. 2V' X 2^". 

(1008) Calculate the dimensions of a steel end journal 
subjected to a load of 17,800 lb. ; direction, variable. The 
bearing pressure is not to exceed 750 lb. per sq. in. of 
projected area. Ans. 3f" X 6^' 

(1009) Calculate the dimensions of a wrought-iron end 
journal sustaining a load of 1,7501b.; direction, constant. 
The pressure per square inch of projected area is not to 
exceed 275 lb. Ans. If" X 4". 

(1010) Calculate the dimensions of a wrought-iron end 
journal, assuming the length to be twice the diameter and 
the bearing pressure not to exceed 550 lb. per sq. in. The 
load on the journal is 3,750 lb. Ans. 1|-" X 3f". 

(1011) Calculate the dimensions of the steel journal of a 
car axle bearing a load of 10,000 lb., which may be consid- 
ered variable in direction on account of the spring of the 
car. Speed of car is 40 miles per hour; diameter of truck 
wheel, 2f ft., and the allowable bearing pressure is 

p = j^ — , where N is the number of revolutions per 

minute. Design and draw the journal. 

(1012) {a) Calculate the dimensions of a wrought-iron 
neck journal, assuming the length equal to 2|- times the 
diameter. The load is 9,600 lb., variable in direction. 
(^) What is the pressure per square inch of projected area ? 

A^, ] (^) ^r X 5^V, say 5^. 
( (b) 860 lb. per sq. in. 

(1013) Find the diameter of a wrought-iron pivot run- 
ning in gun-metal bearings, the load being 25 lb. and the 
number of revolutions 3,000 per minute. Ans. 1^". 

(1014) What should be the diameter of a steel pivot (on 
gun-metal bearings) carrying a load of 4^ tons and making 
65 R. P. M. ? Ans. 4f ". 

(1015) A shaft 6" in diameter receives an end thrust of 
7, 800 lb. on 4 collars. Find the diameter and thickness of col- 
lars. Draw the portion of the shaft with collars. Scale 3" = 1'. 



MACHINE DESIGN. 1727 

(1016) Calculate the diameter of a steel engine shaft 8 ft. 
long carrying a fly-wheel weighing 10 tons; the center of 
the wheel is 3 ft. from the center of one of the bearings. 
The engine develops 300 H. P. at 60 R. P. M., has a stroke 
of 48", and a M. E. P. of 62 lb. per sq. in. Assume 5^ = 
9,000 lb. per sq. in. Ans. 8 J". 

(1017) A line-shaft is required to transmit 90 H. P. at 
110 R. P. M. The average bending moment, due to weights 
of pulleys, belt pull, etc., is about f of the twisting moment. 
Assuming a wrought-iron shaft and a safe stress of 7,000 lb. 
per sq. in., find the diameter of the shaft. Ans. 4^". 

(1018) The inner diameter of a hollow shaft is 6", the outer 
diameter 13". Find the diameter of a solid shaft of the same 
material and having the same strength. Ans. 12^". 

(1019) In designing a propeller shaft, it is found that the 
proper diameter of a solid steel shaft is 16|". It is desired, 
however, to make the shaft hollow. Assuming the diameter of 
the hole to be -J- the diameter of the shaft, find the inside and 
outside diameters of the hollow shaft. . { ^i — I'^i"- 

(1020) Design and draw a universal joint for shafts 1^ 
in diameter. Draw full size. 

(1021) Design and draw a flange coupling for a shaft 3J* 
in diameter. Draw half size. 

(1022) Calculate the following for the solid flange coup- 
ling of a propeller shaft 11" in diameter: (a) Number of 
bolts; (d) diameter of bolts; (c) diameter of bolt circle; 
(d) diameter of flange; (e) thickness of flange. See Fig. 674. 

r (a) 6 bolts. 

(d) 2r. 

Ans. ^ (c) 16Y' 

(d) 22^^ 

(e) 3i", nearly. 

(1023) A wrought-iron pivot runs at a speed of 600 
R. P. M. in gun-metal bearings. The load being 100 lb., 
what should be the diameter ? Ans. 1^ 



MACHINE DESIGN 

(ARTS. 2004-2074.) 



EXAMINATION QUESTIONS. 

(1024) For what purpose are bearings bushed ? 

(1025) For what reason are ball and socket bearings 
used ? 

(1026) A cast-iron gear makes 120 R. P. M. and transmits 
20 H. P. If the pitch diameter is 30", what should be (a) 
the circular pitch, and (I?) the number of teeth ? 

Ans i (^) l-^^^l"- 
' i.{d) 76 teeth. 

(1027) Compute the leading dimensions for a spur gear, 
working under the following conditions: Diameter of pitch 
circle = 4 ft. 8" ; revolutions per minute = 60 ; horsepower 
transmitted = 300. 

(1028) Suppose a friction wheel faced with wood to drive 
another 6 ft. in diameter, at 110 revolutions per minute, the 
force pressing the wheels together being 280 lb. Take / 
equal to -J and calculate the horsepower. Ans. 5.864 H. P. 

(1029) What is meant by counterbalancing a pulley ? 

(1030) (a) What is the diameter of wires composing a I'' 
wire rope containing 42 wires ? {^) What is the weight of 
this rope per foot of length ? 

(1031) What is the safe load for a crane chain, the links 
being forged from l^-'' round iron ? Ans. 10,312.5 lb. 

(1032) When using divided bearings, how should the 
brasses be divided ? 

(1033) Design and draw a pedestal similar to the one 
described in Art. 2014, and shown in Fig. 692, taking d 
equal to 14", and using a scale of 3" = 1 ft. 

(1034) Of what materials are gears usually made ? 

For notice of copyright, see page immediately following the title page. 



1730 MACHINE DESIGN. 

(1035) How many arms should be given a gear having a 
diameter of 56" and a diametral pitch of 1^" ? Ans. 6 arms. 

(1036) For what purposes may friction gearing be used ? 

(1037) When transmitting power by means of a hemp 
rope, what should be the ratio between the diameter of the 
pulley and that of the rope ? 

(1038) What should be the least value for the ratio of 
the diameter of a wire rope pulley to that of the rope ? 

(1039) What are gaskets, and for what are they used ? 

(1040) What are the seats or steps of a bearing, and for 
what purpose are they used ? 

(1041) For what purpose are loose disks used in connection 
with pivot bearings ? 

(1042) Design a pedestal similar to the one described in 
Art. 201 1 . Take d equal to 3^'', and use a scale of 6" = 1 ft. 

(1043) (a) What is a hanger ? (b) A wall bracket ? 

(1044) A cast gear 44" in diameter is to transmit 21 H. P. 
when making 50 R. P. M. What should be the least breadth 
of the teeth ? Ans. 2J". 

(1045) The circular pitch of a 24" worm wheel is .5236"; 
if the worm has 4 threads, what is the efficiency ? Take 
diameter of worm as 2^. Ans. 62.62^. 

(1046) (a) Of what materials are belts made ? (/^) For 
what situations are rubber belts preferred ? 

(1047) What should be the least thickness of the rim 
of a crowning belt pulley having a diameter of 52" and a 
breadth of 10" ? Ans. |". 

(1048) The distance between two wire rope pulleys is 
420 ft. What will be the deflection of the rope midway 
between the pulleys, if the rope is ^" in diameter, is com- 
posed of 42 wires, and the tension due to its weight is 
2,023 lb.? Ans. 12 ft. 

(1049) What is considered to be the greatest safe rim 
speed of a cast-iron pulley ? 



MACHINE DESIGN. 1731 

(1050) What are the advantages claimed for rope belting ? 

(1051) What horsepower may be transmitted by a rope 
wheel 18 ft. in diameter which is grooved for 24 turns of 
lY rope ? R. P. M. = 72. Ans. 484.32 H. P. 

(1052) The deflection of a 1^" wire rope having 42 wires 
is 9 ft. 3"; the distance between the pulleys being 386 ft., 
what is the tension due to the weight of the rope ? 

Ans. 4,511 lb. 

(1053) Calculate the dimensions of the plate link gearing 
chain for a working load of 5,600 lb. 

(1054) Design a wall bracket similar to the one described 
in Art. 2019. Take d equal to 2|-", and use a scale of 
6" = 1 ft. 

Note. — Pay particular attention to the last sentence of Art. 2019. 

(1055) A gear-wheel 40" in diameter has 6 arms and a 
diametral pitch of 1^. The breadth of the teeth being 6'\ 
find the dimensions of the arms. Consider the arms as 
being elliptical in cross-section. 

(1056) Power is transmitted by a f" iron rope having 
42 wires. Find (a) the horsepower transmitted when the 
speed of the rope is 5,000 ft. per min., distance between 
pulleys is 425 ft., and the diameter of the pulleys is 13 ft. 
{b) Find the greatest deflection in the rope. 

Ans. \ (f ) IS^ H- P- 
( (l?) 14.7 ft. 

(1057) How many arms should be given to a 21" gear- 
wheel having 84 teeth ? Ans. 5 arms. 

(1058) Calculate the size of the tapering arms of a 4-arm 
pulley, 22" in diameter, and with an 8" face; a double belt 
is to be used. 

(1059) The hub of a split pulley is 7" long and If" thick. 
If the hub is held by 6 bolts, what should be the diameter 
of the bolts ? Ans. J". 



1732 MACHINE DESIGN. 

(1060) Design the following and draw on one plate : 

1. A belt pulley similar to the one shown in Fig. 
720. Take diameter equal to 48", face 12", 
and bore 2^. Use a scale of 3" = 1 ft. 

2. A foot-step bearing similar to that shown in 
Fig. 698. Take d equal to 3", and use a scale 
of 6" = 1 ft. 

3. A crane hook similar to that shown in Fig. 735, 
except that the swivel end shown at A should 
be used instead of the eye. Take d equal to 
2J", and use a scale of 3" = 1 ft. 

4. A pipe flange as shown in Fig. 736. Take a 

equal to 12", and use a scale of lY = 1 ft. 

Note. — The dots in Table 53 indicate that the last dimension above 
them should be used. 

(1061) What is the weight of 875 ft. of If" manila rope ? 

Ans. 803.9 lb. 

(1062) {a) Calculate the horsepower which can be trans- 
mitted by a If" cotton rope at a velocity of 3,800 ft. per 
minute. (<^) What is the horsepower as estimated from the 
diagram, Fig. 723? Ans. (a) 38.223 H. P. 

(1063) (a) What should be the diameter of the wires 
composing an iron wire rope transmitting 125 H. P. at a 
velocity of 4,900 ft. per minute? (d) What should be the 
diameter of the pulley ? {c) Of the rope ? The rope is to 
contain 42 wires. { (a) .0694". 

Ans. \ (3) 9 ft. 10*. 



k 



I 



A KEY 

TO ALL THE 

QUESTIONS AND EXAMPLES 

CONTAINED IN THE 

EXAMIIS^ATIOX QUESTIOIS^S 

Included in this Volume. . 



The Keys that follow have been divided into sections cor- 
responding to the Examination Questions to Avhich they 
refer. The answers and solutions have been numbered to 
correspond with the questions. When the answer to a ques- 
tion involves a repetition of statements given in the Instruc- 
tion Paper, the reader has been referred to a numbered 
article, the reading of which will enable him to answer the 
question himself. 

To. be of the greatest benefit, the Keys should be used 
sparingly. They should be used much in the same manner 
as a pupil would go to a teacher for instruction with regard 
to answering some example he was unable to solve. If used 
in this manner, the Keys Avill be of great help and assist- 
ance to the student, and will be a source of encouragement 
to him in studying the various papers composing the Course. 



I 



ELEMENTARY MECHANICS. 

(QUESTIONS 355-453.) 



(355) Use formulas 18 and 8. 

Time it would take the ball to fall to the ground = ^ = 



2A_^2 X 5.5 



= .58484 sec. 



g ' 32.16 

The space passed through by a body having a velocity of 
500 ft. per sec. in .58484 of a second = 5 = F/ = 500 X 
.58484 = 292.42 ft. Ans. 

(356) Use formula 7. 

■ffX 3.1416 X 160 _ ^^. . 

i^ — = 5o. 85 ft. per sec. Ans. 

(357) 160 -J- 60 X 7 = %: revolution in \ sec. 360° X 

/vl 7 

A = 137I - 1370 8' 34^ . Ans. 

(358) (a) See Fig. 20. 36^ = 3'. 4 -^ 3 = | = number 

o 
of revolutions of pulley to one revolution of fly-wheel. 

4 
54 X ^ = 72 revolutions of pulley and drum per min. 100 -r- 
o 

18 \ 

— X 3.1416) = 21.22 revolutions of drum to raise elevator 

21 22 
100 ft. ^^—^ X 60 = 17.68 sec. to travel 100 ft. Ans. 

{b) 21.22 : .r :: 30 : 60, or jr = ^f^ = 42.44 rev. 

For notice of the copyright, see page Immediately following the title page. 



( 



166 ELEMENTARY MECHANICS. 

per min. of drum. The diameter of the pulley divided by 




Fig, 20. 

3 

the diameter of the fly-wheel = — , which multiplied by 

42.44 = 31.83 revolutions per min. of fly-wheel. Ans. 

(359) See Arts. 857 and 859. 

(360) See Art. 861. 

(361) See Arts. 843 and 871. 

(362) See Art. 871. 

(363) See Arts. 842, 886, 887, etc. 

The relative weight of a body is found by comparing it 
with a given standard by means of the balance. The abso- 
lute weight is found by noting the pull which the body will 
exert on a spring balance. 

The absolute weight increases and decreases according to 
the laws of weight given in Art. 890 ; the relative weight is 
always the same. 

(364) See Art. 861. 

(365) See Art. 857. 



I 



ELEMENTARY MECHANICS. 167 

(366) See Art. 857. 

(367) If the mountain is at the same height above, and 
the valley at the same depth below sea-level respectively, 
it will weigh more at the bottom of the valley. 

(368) —^-^ = 6 miles. Using formula 12,d':R':: 

W: w,weh2iYew — —j^ = -^ 3"966^ = 19,939.53 + lb. 

= 19,939 lb. 8-oz. Ans. 

(369) Using formula 11, J^ : d :: W : w, we have 

^^^^ 3,958X20,000 ^^ 3^3^ lb. = 19,989 lb. l^ 
A. 3,960 4 

oz. Ans. 

(370) See Art. 870. 

(371) See Art. 894. 

(372) The velocity which a body may have at the in- 
stant the time begins to be reckoned. 

(373) Because the man after jumping tends to continue 
in motion with the same velocity as the train, and the sud- 
den stoppage by the earth causes a shock, the severity of 
which varies with the velocity of the train. 

(374) See Arts. 870 and 871. 

(375) See Art. 872. 

(376) That force which will produce the same final 
effect upon a body as all the other forces acting together is 
called the resultant. 

(377) {a) If a 5-in. line =^ 20 lb., a 1-in. line = 4 lb. 
1-^4 = - in. = 1 lb. Ans. (d) 6^ -^ 4 = 1.5625 in. = 

6- lb. Ans. 
4 

(378) Those forces by which a given force may be 
I). 0. III.— 30 



168 



ELEMENTARY MECHANICS. 



replaced, and which will produce the same effect upon a 
body. 

(379) Southeast, in the direction of the diagonal of a 
East . square. See Fig. 21. 




(380) 4' 6" = 54". 



54x2XtX 
4 



Fig. 21. 



tance from c equal to 
Art. 911. 



.261 = 21.141 lb. = weight of lever. 
Center of gravity of lever is in the 
middle, at a, Fig. 22, 27" from each 
end. Consider that the lever has no 
weight. The center of gravity of 
the two weights is at by at a dis- 
47 X 54 



47 + 'J'l 



= 21. 508" = bc. Formula 20, 



54.'- 



5: 



t 



-27'! 



\^ , ( \ 



J 



>**- 




b\ 



-21.S0S- 




PlG. 22. 

Consider both weights as concentrated at b^ that is, 
imagine both weights removed and replaced by the dotted 
weight W, equal to 71 + 47 = 118 lb. Consider the weight 
of the bar as concentrated at a, that is, as if replaced by 
a weight w = 21.141 lb. Then, the distance of the balancing 

. ^ r ^ 21.14:1 X 5.492 _„., . 

pomt /, from e, or fe, = 21.141 -|- 118 = -^^^ ' ^'^^^ ^^ = 

27 - 21.508 = 5.492". Finally,/^+ ^// =//i = .834 + 21.508 
= 22.342" = the short arm, Ans. 54-22.342 = 31.658" = 
long arm. Ans. 



ELEMENTARY MECHANICS. 
(381) See Fig. 23 




169 



7Slh. 




(382) See Fig. 24. 




nsih. 




rsTb. ^' 
Fig. 24. 



(383) 46 — 27 = 19 lb., acting in the direction of the 
force of 46 lb. Ans. 



170 



ELEMENTARY MECHANICS. 



(384) {a) 18 X 60 X 60 = 64,800 miles per hour Ans 
(d) 64,800 X 24 = 1,555,200 miles. Ans. 




Fig. 25. 



ELEMENTARY MECHANICS. 



171 



(385) (a) 15 miles per hour = — — - — '—-- = 22 ft. per 

oO X oO 

sec. As the other body is moving 11 ft. per sec, the 
distance between the two bodies in one second will be 
22 + 11 = 33 ft., and in 8 minutes the distance between 
them will be 33 X 60 X 8 = 15,840 ft., which, divided by the 



Ans. 



number of feet in one mile, gives ^ ' ^^ = 3 miles. 

5,280 

(l?) As the distance between the two bodies increases 

33 ft. per sec, then, 825 divided by 33 must be the time 

825 
required for the bodies to be 825 ft. apart, or -— - = 25 

sec. Ans. 

(386) See Fig. 25. 

(387) (a) Although not so stated, the velocity is 
evidently considered with reference to a point on the shore. 
10 — 4 = 6 miles an hour. Ans. 

(d) 10 + 4 = 14 miles an hour. Ans. 

(c) 10 — 4 + 3 = 9, and 10 + 4+ 3 = 17 miles an hour. Ans. 

(388) See Fig. 26. 




Fig. 26. 



172 



ELEMENTARY MECHANICS. 



(389) See Fig. 27. By rules 2 and 4, Art. 754, bc = 
87 sin 23° = 87 X .39073 = 33.994 lb., ac^m cos 23° = 
87 X .92050 = 80.084 lb. 




SOlh, 



Fig. 27. 



(390) See Fig. 28. {b) By rules 2 and 4, Art. 754, 
hc^ 325 sin 15° = 325 X .25882 = 84.12 lb. Ans. 

(a) ^^ = 325 cos 15° = 325 X. 96593 =313.93 lb. Ans. 




Fig. 28. 



(391) Use formula lO. 

W 125 

tn •=. — = 

g 32.16 



= 3.8868. Ans. 



ELEMENTARY MECHANICS. 173 

(392) Using formula 10,^ = — , lV=m£-=53.7 X 
32.16 = 1,727 lb. Ans. 

(393) (^) Yes. (d) 25. (c) 25. Ahs. 

(394) (a) Using formula 12^ d' : R' :: IV : w, d = 

,/WW ^ / 4,000' X 141 ^ ^^^_^3g ^.j^^_ ^^^_^3g _ 

2£/ 100 

4,000 = 749.736 miles. Ans. 

R iv 
{b) Using formula 11, R \ d :\ W : w, d = -^ = 

4,000 X 100 _ 2 g36 88 miles. 4,000 - 2,836.88 = 1,163.12 
141 > > » 

miles. Ans, 

(395) {a) Use formula 18, 

= 18.12 sec. Ans. 



2A_./2X 5,280 



g ' 32.16 
{b) Use formula 13,^ = ^/ = 32. 16X18. 12 = 582. 74 ft. per 

sec, or, by formula 16, ^ = V^g^ = V^ X 32.16 X 5,280 = 
582.76 ft. per sec. Ans. 

The slight difference in the two velocities is caused by not 
calculating the time to a sufficient number of decimal places, 
the actual value for / being 18.12065 sec. 

(396) Use formula 25. Kinetic energy = Wk = — — . 
Wk = 160 X 5,280 = 844,800 ft. -lb. 

Wv' 160X582.76' Q,,,voaf^iK a 
-27" = 2X32.16 = ^^^'^^^ ^''-^^' ^^^- 

(397) (a) Using formulas 15 and 14, 

/i= ^ = ^ '^^,„ = 86,592 ft. = 16.4 miles. Ans. {b) 
2^ 2 X 32.16 ' ^ ^ 



V 



f = —= time required to go up or fall back. Hence, total 

o 

time = — sec. = — ^—r^ = 2.4461 mm. = 2 mm. 26.77 

g 60 X 32.16 

sec. Ans. 



in ELEMENTARY MECHANlCi^. 

(398) 1 hour =: GO min., 1 day = 24 hours; hence, 1 

day = 60 X 24 = 1,440 min. Using formula 7, V — ~', 

, ^, 8,000 X 3.141G _ __ , „ . . 

whence, V = t—ttt. = 17.453 + miles per mm. Ans. 

' 1,440 ^ 

(399) (a) Use formula 25. 

T^. ,. W^^^' 400 X 1,875 X 1,875 _ . __ .„ _ 
Kmetic energy = — — = • ^ ^ g,^ .^^ = 21, 863, 339. 55 

ft. -lb. Ans. 

(P) ^^'^f^nf'^^ = 1^' 9^1-6^ ft. -tons. Ans. 

/v, uuu 

(^) See Art. 961. 

Striking force X ^r, = 21,863,339.55 ft. -lb., 

or striking force = ^^'^^^^^'^^ == ^^' ^^^' ^^^ lb. Ans. 

T2 

2 X 200 



(400) Using formula 18, t^ yVi — ^l 



g ' 32.16 
3.52673 sec, when^= 32.16. 



t = i/AKl22 = 4. 47214 sec. , when g = 20. 
'^ 20 

4. 47214 - 3. 52673 = 0. 94541 sec. Ans, 

(401) See Art. 910. 

(402) See Art. 963. 

(403) {a) See Art. 962. 

^ in W 800 ^ T^ W 500 

1)=-- ^ — • z/= . Hence, i^ = — = 



V gv 1,728' ' gv 32.16XtV% 

33.582. Ans. {b) In Art. 962, the density of water was 
found to be 1.941. {c) In Art. 963, it is stated that the 
specific gravity of a body is the ratio of its density to the 

density of water. Hence, -^r-^rrj' = 17.3 = specific gravity. 

If the weight of water be taken as 62.5 lb. per cu. ft., the 
specific gravity will be found to be 17.28. Ans. 

(404) Assuming that it started from a state of rest, 
formula 1 3 gives z/ = ^ / = 32. 16 x 5 = 160. 8 ft. per sec. 



ELEMENTARY MECHANICS. 175 

(405) Use formulas 17 and 13- >^=i-^/' = 

—-— X 3' = 144.72 ft., distance fallen at the end of third 

second. 

v=£-t= 32.16 X 3 = 96.48 ft. per sec, velocity at end of 
third second. 

96.48 X 6 = 578.88 ft., distance fallen during the remain- 
ing 6 seconds. 

144.72 + 578.88 = 723.6 ft. = total distance. Ans. 

(406) See Art. 961. 

Striking force x r^ = 8 X 8 = 64. Therefore, striking 

r A. 
force = — = 1,536 tons. Ans. 
X 
12 

(407) See Arts. 901 and 902. 

(408) Use formula 19. 

Centrifugal force = tension of string = .00034 WRN* = 

.00034 X (.5236 X 4' X .261) X t^ X 60' = 13.38-}- lb. Ans. 

(409) (80' - 70') X .7854 X 26 X .261 -5- 2 = weight of 
J of rim. 

__ 80-10 _ 35 ^ 

^ " 2 X 12 - 12 ^' 
According to Art. 904, F= .00034 IV RN^ -~ 3.1416 = 
.00034 X (80' - ^0') X .7854 X 26 X -261 ^ 35 ^ ^^^, ^ 

3 1416 = 38,641 lb. Ans. 

(410) (^) Use formulas 11 and 12, R : d 'i W iw^ 

„. wR 1x4,000 .... . 
ox W — —J- — — TTT^T — = 40 lb. Ans. 
a 100 

(b) d' :R' :;W : w,orw = ^^^?Tnn, - = 38.072 lb. Ans 
^ ' * 4,100' 

(411) See Art. 955. 



176 ELEMENTARY MECHANICS. 

(412) Use formula 12. 



d^ iR'iiW: w, ox d 



3 

T6" 



13,064 mi., nearly. 



13,064 — 4,000 = 9,064 miles. Ans. 



2 X 50 



(413) Use formula 18. t = \/^ = ^-^^^ 

^32.16 

1, 7634 sec., nearly. 



1.7634 X 140 = 246.876 ft. Ans. 
(414) WR — -^ sec. Use formula 17. 



^ = 1^.-1 X 32.16 xg) 



1.78| ft. =1 ft. 9.44 in. 

Ans. 




(416) See Arts. 906, 907. 
(41 6) See Arts. 908, 909. 



ELEMENTARY MECHANICS. 17? 

(417) No. It can only be counteracted by another 
equal couple which tends to revolve the body in an opposite 
direction. 

(418) See Art. 914. 

(419) Draw the quadrilateral as shown in "Fig. 29. 
Divide it into two triangles by the diagonal B D. The 
center of gravity of the triangle B C D \s found to be at ^, 
and the center of gravity of the triangle A B D is found to 
be at d (Art. 914). Join a and d by the line a b, which, on 
being measured, is found to have a length of 4.27 inches. 
From C and A drop the perpendiculars C F and A G q^h the 

diagonal B D, Then, area of the triangle A B D — - {AG X 

/V 

B D), and area of the triangle B C D=\{C F y^ B U). 

Measuring these distances, B D^^\\\ C F=6.1% and 
A G=^ 7.7". 

Area A B D =\ x 7.7 X 11 = 42.35 sq. in. 

Kx^d.BCD-\ X 5.1 X 11 = 28.05 sq. in. 

According to formula 20, the distance of (7, the center of 

,^ . ,. 28.05 X 4.27 -, .v o^v, ^ .;, 

gravity, from b is ^g 95 1 42 35 ~ Therefore, the cen- 

ter of gravity is on the line ^ ^ at a distance of 1.7" from b, 

(420) See Fig. 30. The 
center of gravity lies at the 
geometrical center of the penta- 
gon, which may be found as 
follows: From any vertex draw 
a line to the middle point of the 
opposite side. Repeat the 
operation for any other vertex, 
and the intersection of the two 
lines will be the desired center 
of gravity. 




i-lG. 50. 



I 



178 



ELEMENTARY MECHANICS. 



(421) See Fig. 31. Since any number of quadrilaterals 
can be drawn with the sides given, any number of answers 
can be obtained. 

Draw a quadrilateral, the lengths of whose sides are equal 
to the distances between the weights, and locate a weight on 
each corner. Apply formula 20 to find the distance C^ W^ ; 



thus, C W. = 



9 X 18 
9 + 21 



= 5.4". Measure the distance C^ W^\ 




Fig. 31. 

suppose it equals say 36''. Apply the formula again. 
15 X 36 



C C = 
' ' 15 +(9 + 21) 



31.7". 

Apply the formula again. C^C = 



= 12". Measure C^ W^ ; it equals say 

17 X 31.7 



17 + 15 + 9 + 21 
C is center of gravity of the combination. 



8.7". 



(422) 'Let A B C D E, Fig. 32, be the outline, the 
right-angled triangle cut-off being E S D. Divide the 
figure into two parts by the line m n^ which is so drawn 
that it cuts off an isosceles right-angled triangle m B n^ 
equal in area to E S D^ from the opposite corner of the 
square- 



ELEMENTARY MECHANICS. 



179 



The center of gravity of A inn C D E is then at C^ , its 
geometrical center. Bin =-4: in. ; angle Binr= 45°; there- 
A m B 




fore, ^r=:^'M X sin^7;^r = 4 X .707 = 2.828 in. C^, the 

2 
center of gravity of Bmn^ lies on Br^ and B C^^= —Br 

o 

- I X 2.828 = 1.885 in. ^ (^T, = yi ^ X sin ^^ (T, = 14 X 

1 

-- 1.885 = 8.013 in 

Area 
4x4 



sin 45° = 14 X .707 = 9.898 in. C^C, = B C^- B C^ = d.SdS 



Area ABODE- 14^-^-4-^=188 sq. in. Area mBn 



2 



2 
sq. in. 



8 sq. in. Area AmnCDE-V^^-'^^X'^^ 



The center of gravity of the combined area lies at C^ at 



180 



ELEMENTARY MECHANICS. 



a distance from C^, according to formula 20 (Art. 911), 

. ^ SX C^C^ 8 X 8.013 _.. . ^ ^ _.. . 
"^^^^ '^ "180^ = -18^ = '^^^ ^^- ^^ ^ = '^^^ ^^• 
^ (:- = ,5 (7, - {7, <:: = 9. 898 - . 341 = 9. 557 inches. Ans. 

(423) (^) In one revolution the power will have moved 

through a distance of 2 X 15 X 3.1416 = 94.248", and the 

1" 
weight will have been lifted j . The velocity ratio is then 



94.248 -^ 7- 
4 



376.992. 

376. 992 X 25 = 9, 424. 8 lb. Ans. 
(a) 9,424.8 - 5,000 = 4,424.8 lb. Ans. 

(c) 4,424.8 -^ 9,424.8 = 46.95^. Ans. 

(424) See Arts. 920 and 922. 

(425) Construct the prism ABED, Fig. 33. From 
G, draw the line £ F. Find the center of gravity of the 

A/ 




rectangle, which is at C^, and that of the triangle, which is 



ELEMENTARY MECHANICS. 181 

at C^. Connect these centers of gravity by the straight line 
C^ C^ and find the common center of gravity of the body by 
the rule to be at C. Having found this center, draw the 
line of direction C G. If this line falls within the base, the 
body will stand, and if it falls without, it will fall. 

(426) {a) 5 ft. 6 in. = 66". 66 -^ 6 - 11 = velocity 
ratio. Ans. 

{b) 5 X 11 = 55 lb. Ans. 

(427) 55 X .65 = 35.75 lb. Ans. 

(428) Apply formula 20. 5 ft. = 60^ ~t^ = 

ioU -f- DO 

9.7674 in., nearly, = distance from the large weight. Ans. 

(429) {a) 1,000-^50 = 20, velocity ratio. Ans. See 
Art. 945. {b) 10 fixed and 10 movable. Ans. {c) 50 ^ 95 
= 52.63^. Ans. 

(430) PX circumference = ^X q, or 60 X 40 X 3.1416 
= Wx\yOr IV = 60 X 40 X 3.1416 X 8 = 60,318.72 lb. Since 

o 

the efficiency of combination is 40^, the tension on the stud 
would be .40 X 60,318.72 = 24,127.488 lb. Ans. 



(431) (a) 4/20'+ 5^* = 20.616 ft. = length of inclined 
plane. 

P X length of plane = W X height, or P X 20.616 = 
1,580 X 5. 

p_ L,ooKj A ^ __3g3 2 ii3^ Ans. [b) In the second case, 
20.616 
P X length of base =Wx height, or /'x 20 = 1,580 X 5; 

hence, P= ^-^y^ = 395 lb. Ans. 

(432) ^rx2 = 42x 6, or PF=^^-^=126 lb. 

1 68 
126 + 42 = 168 lb. 168 x 1 = ^' X 12, or IF' = — = 14 lb. 

Ans. 



182 



ELEMENTARY MECHANICS. 



(433) See Fig. 34. Px 14 X 21 X 19 = 2^ X 3^ X 2^ 

/v 4 o 

X 725, or 

j^_ 2i X 3i X 2| X 72 5 . .,^ ,, ._ 
^- 14X21X19 ~^^-^^^^^- ^^^• 






.-SOfi—. 



p 

Fig. 34. 



^ 



i^ 






(434) See Fig. 35. {a) 35 X 15 X 12 X 20 = 5 X 3 i- X 

3 X ^, or 

„. 35 X 15 X 12 X 20 . ... ,, . 
^= 5x3ix3 = ^'400 lb. Ans. 




Fio. 86. 



ELEMENTARY MECHANICS. 183 

4 
(d) 2,400 -^ 35 = G8y = velocity ratio. Ans. 

(<:) 1,932 ^2,400 =.805 = 80. 5^. Ans. 

(435) In Fig. 36, let the 12-lb. weight be placed at A, 
the 18-lb. weight at B, and the 15-lb. weight at B, 
Use formula 20. 

^^ ^ ^^ = 6" = distance CB= distance of center of 
18+12 




Fig. 36. 



gravity of the 12 and 18-lb. weights from B. Drawing C\ D, 

r r — 1^ X ^1 -^ —^CD Measuring the distances 

^i^-(12 + l8) + 15 3 ^ 

of C from BD, DA,d.n6iAB, it is found that Ca = 3.45", 

(7^ = 5.25", and {7^=4.4". Ans. 

(436) {a) Potential energy equals the work which the 
body would do in fallins: to the ground = 500 X 75 = 
37,500 ft. -lb. Ans. 
D. 0. III.— SI 



184 



ELEMENTARY MECHANICS. 



(d) Using formula 18; 






2X 75 
32.16 



= 2.1597 



sec. = .035995 min., the time of falling. 
37,500 



— = 31.57 H. P. Ans. 
33,000 X .035995 

(437) 127 -f- 62.5 = 2.032 = specific gravity. Ans. 
62.5 



(438) 



1,728 



X 9. 823 = .35529 lb. Ans. 



(439) Use formula 21. ^=(|^), 
or ^ -^/^ ^ ^-^^ X .48 = 499.2 lb. Ans. See j^ 
Fig. 37. 



or 



(440) See Art. 961. 
^.. /3 . .A_lVv'_ 1.5X25- 
^^ V8 / ^£- ~2X 32.16' 

1.5 X 25^^ 

/r=!><iH^ = 466.42 lb. Ans. 

(441 ) (a) 2,000 -^ 4 = 500 = wt. of cu. 

ft. 500 -J- 62. 5 = 8 = specific gravity. Ans. 



500 




= .28935 lb. Ans. 




Fig. 37. 



Fig. 38. 



(442) See Fig. 38. 14.5x2 
= 29. 30 X 29 = Wx 5, or W= 

22_>il^ = 174 lb. Ans. 
5 

(443) 75 X .21 = 15.75 lb. 

Ans. 



(444) (a) 900 x 150 = 135,000 ft. -lb. Ans. 
135,000 



15 



= 9,000 ft. -lb. per min. Ans. 



,,. 9,000 3 -J j3 . 

(^) 3^000 ^n^-^- ^^^' 



ELEMENTARY MECHANICS. 



185 



(445) 900 X .18 X 2 = 324 lb. = force required to over- 
come the friction. 900 -}- 324 = 1,224 lb. = total force. 

M!i^ = .37091H.P. Ans. 
15 X 33,000 

(446) 18-^88 = .2045. Ans. 

(447) See Art. 962. i.= i|=^J^ 

(448) See Fig. 39. 125 
-47.5 = 77.5 lb. = down- 
ward pressure. 

77.5 --4 = 19.375 lb. 
s= pressure on each support. 

Ans. 

(449) See Fig. 40. 



= 12.438. 

Ans. 




Fig. 89. 




Fig. 40. 



(450) See Fig. 41. 4.5 -^ 2 = 2.25. 
12 



2.25 



X 6 X 30 = 960 lb. Ans. 



(451) (^) 960 -^ 30 = 32. Ans. 

(3) 790 -r- 960 = .829.«.-= 82.29<. Ansk 



186 



ELEMENTARY MECHANICS. 



;452) (a) See Fig. 42. 475 + (475 X .24) = 589 lb. 

Ans. 
{d) 475-5-589 = .8064 = 80. 64^. Ans. 





Fig. 41. Fig. 42. 

(453) (a) By formula 23, U= FS=: Qx25 = 150 
foot-pounds. Ans. 

(5) 2lsec. = |-=l^min. 

Using formula 24, Power = -=^ = ^= 3,600 ft. -lb. per 
mm. Ans. 



STRENGTH OF MATERIALS, 

(QUESTIONS G89-748.) 



(689) See Arts. 1336, 1339, and 1338. 

(690) See Arts. 1344, 1345, 1352, and 1354. 

(691) vSee Art. 1347. 

(692) Use formula llO. 

ir = ^;therefore,. = ^. 

A e AE 

^ = .7854X 2^ /=rlO X 12; /^=: 40 X 2,000; iS'rr 25,000,000. 

Therefore . - 40 X 2,000 X 10 X 12 _ , 

1 neretore, e - ^ ^^^^ ^ ^ ^ 25,000,000 " ' ^^^^^ ' ^''^' 

(693) Using formula 1 lO, 

p/ 7 000 V 7-1- 

^ = Z-. = .7854 X a? X .009 = ^9.^08,853.2 lb. per sq. in. 

^ Ans. 

(694) Using formula 1 lO, 

r^ PI ^ AeE Ux 2 X .006 X 15,000,000 
^=-^,orP=-^ = -^ ^^02 = 

2,500 lb. Ans. 

(695) By formula 110, 

^ Z'/ , AeE .7854 X 3'X .05 X 1,500,000 ^^^ ^^, 

^ = Z7'^'^ = -p-=^ 2;000 =265.07. 

Ans. 

(696) Using a factor of safety of 4 (see Table 28), 
formula 108 becomes 

For notice of copyright, see page immediately following the title pa^e. 



254 STRENGTH OF MATERIALS. 



p= AS, ^^^^ iP 4X ex 2000 ^ 3^^^^^^ .^ 
4 vS 55,000 , ^ 



(697) From Table 23, the weight of a piece of cast iron 
1" square and 1 ft. long is 3.125 lb.; hence, each foot of 
length of the bar makes a load of 3.125 lb. per sq. in. The 
breaking load — that is, the ultimate tensile strength — is 
20,000 lb. per sq. in. Hence, the length required to break 

the Dar is ^^ = G,400 ft. Ans. 

(698) Let / = the thickness of the bolt head; 

d= diameter of bolt. 
Area subject to shear =.7: d t. 

Area subjected to tension = --n d"^, 

S^ = 55,000. ^'3 = 50,000. 
Then, in order that the bolt shall be equally strong in both 

tension and shear, 7t d t S^ = —tz d^ S,, 

or/ = ^ = g-. = |2^^0 = .20C'. Ans. 
4:7rdS^ 4^3 4 X 50,000 

(699) Using a factor of safety of 15 for brick, formula 
108 gives 

15 ' 
A = {2iX Si) sq. ft. = 30 X 42 =r 1,260 sq. in. ; S^ = 2,500. 

Therefore, P= ^^^^^ X ^^^QQ _ 210,000 lb. = 105 tons. Ans. 

(700) The horizontal component of the force Pis Pcos 
30° = 3.500 X .860 = 3,031 lb. The area A is 4: a, the ulti- 
mate shearing strength, 5,, 600 lb., and the factor of safety, 8. 



STRENGTH OF MATERIALS. 



265 



Hence, from formula 108, 



P = 



AS^ __ ^aS^ _ aS^ _ 2P_ 2 X 3,031 _ 



a 



S. 



600 



= 10.1^ Ans. 



(701) See Art. 1366. 




8 S' 

Scale of forces 1^=1600 lb. 
Scale of distance 1=32^ 

Fig. 58. 

(702) Using formula 111, with the factor of safety 4 



^d 



2fS, fS, 



or 



2/^ 
5, 



2 X 120 X 48 
55,000 



256 STRENGTH OP MATERIALS. 

Since 40^ of the plate is removed by the rivet holes, 60^ 
remains, and the actual thickness required is 

t _ 2 X 120 X 48 _ 
."GO- .60X55,000 --^^^ ■ ^"'- 
(703) Using a factor of safety of 6, in formula 1 1 1» 

2/5, tS, 



/^ = 



6 



Hence / - ^ - ^ >< ^ >< ^^^ - 18" Ans 
Hence, ^ - 5^ - ^0,000 "'^^ * ^'''- 

(704) Using formula 114, with a factor of safety of 10, 

MOOW!::^ 960,000^. 
■^ 10/<^ /<^ 



Hence, t ^'V^JT =V''' ^ '' ^ '' ^ ' - -^^^^^ 

'^ 960,000 '^ 960,000 ^^^g^ 

(705) From formula 113, 

., -^^ ^,.- /^ ., 2,000X1- _ 4,000 _ 

^~ r + /''''^^-5-/-2,800-2,000~ 800 -^' ^''^• 

(706) See Fig. 58. (a) Upon the load line, the loads 
(9, i, 1-2^ and 2-3 are laid off equal, respectively, to 
F^^ 7^3, and F^ ; the pole P is chosen, and the rays drawn in 
the usual manner; the pole distance 11= 2,000 lb. The 
equilibrium polygon is constructed by drawing a c, c d^ d e, 
and ^/parallel to P O^ Pi, P 2, and P 3, respectively, and 
finally drawing the closing line/ a to the starting point a. 
P m is drawn parallel to the latter line, dividing the load 
line into the reactions in O = P^, and 3 m = R^. The shear 
axis m n is drawn through m, and the shear diagram 
O h I . , . , s' n nt O \^ constructed in the usual manner. To 
the scale of forces m O = 1,440 lb., and 3 m = 2,160 lb. To 
the scale of distances the maximum vertical intercept j/ = 
d'd=dl.^ ft., which, multiplied by i7, = 31.2 X 2,000 = 
62,400 ft. -lb. = 748,800 in.-lb. Ans. 

(^) The shear at a point 30 ft. from the left support = 
Om= 1,440 lb. Ans. 

{c) The maximum shear z=ns' = — 2,160 lb. Ans. 



STRENGTH OF MATERIALS. 



257 



(707) See Fig. 59. Draw the force polygon Ol23Jf5 
in the usual manner, O 1 being equal to and parallel to 
F^, 1-2 equal to and parallel to F^, etc. (9 5 is the resultant. 

Scale of forces 1—40 Ib^ 



Scale of distance 1=2" 



m^^ 



P^ 




Fig. 59. 



Choose the pole P, and draw the rays P O^ P 1, P 2, etc. 
Choose any point, a on F^^ and draw through it a line 
parallel to the ray P 1. From the intersection b of this line 
with F^^ draw a line parallel to P 2; from the intersection 
c of the latter line with F^ produced draw a parallel to 
P 3, intersecting F^ produced in d. Finally, through d, 
draw a line parallel to P ^, intersecting F^ produced in e. 
Now, through a draw a line parallel to P O^ and througli 
e a line parallel to P 5\ their intersection /" is a point on the 
resultant. Through f draw the resultant R parallel to 
O 5. It will be found by measurement that R = 65 lb., that 
it makes an angle of 22^° with 7n 7i, and intersects it at a 
distance of 1^^ from the point of intersection of F^ and 7/i n. 

(708) See Fig. 60. The construction is entirely similar 
to those given in the text. O i, 1-2, and 2-3 are laid off to 
represent -F^, F^^ and F^; the pole Pis chosen, and the rays 



258 



STRENGTH OF MATERIALS. 



drawn. Parallel to the rays are drawn the lines cf the 
equilibrium polygon a b c d g a. The closing line ^ <^ is 
found to be parallel to P 1. Consequently, (9 i is the left 
reaction and 1-3 the right reaction, the former being 6 tons 
Ft Fz F3 

elrons, 2y^Tons, liTon, 




Scale of forces 1=5 tons. 

Scale of distance 1-5' 

Fig. 60. 

and the latter 3 tons. The shear diagram is drawn in the 
usual manner ; it has the peculiarity of being zero between 
F^ and F^. 

(709) The maximum moment occurs when the shear 
line crosses the shear axis. In the present case the shear 
line and shear axis coincide with s /, between F^ and F^\ 
hence, the bending moment is the same (and maximum) at 
F^ and F^^ and at all points between. This is seen to be true 
from the diagram, since k h and b c are parallel. Ans. 

{b) By measurement, the moment is found to be 24 X 12 
= 288 inch-tons. Ans. 

(c) 288 X 2,000 — 576,000 inch-pounds. Ans. 



STRENGTH OF MATERIALS. 



259 



(710) See Arts. 1375 to 1379. 

(711) See Fig. 61. The force polygon O 1 2 3 J^ O is, 
drawn as in Fig. 59, O .^ being the resultant. The equilibrium 
polygon ab c d ga g^^^^ of forces 1^50 lb. 

is then drawn, the scale of distance iSa" 

point ^ lying on the 
resultant. The re- 
sultant R is drawn 
through ^, parallel to 
and equal to O 4^. A 
line is drawn through 
C, parallel to R. 
Through ^ the lines 
g" e and ^/"are drawn 
parallel, respectively, 
to P O and P 4, and 
intersecting the par- 
allel to R, through C 
in e and /; then, e/ 
is the intercept, and 
P u, perpendicular to 
O Ut is the pole dis- 
tance. Pu = ^?>\h.\ 
e f— 1.32". Hence, 
the resultant mo- 
ment is 33 X 1.32 =:= 43.0 in. -lb 

(712) The maximum bending moment, M~ W-^,-^ (see 




Fig. 6 of table of Bending Moments) = 4 X 2,000 X 



14 X 8 

22 



40,727i-\ ft. -lb. =488,727 in.-lb. 
mula 117, 



Then, according to for- 



/^ 



= 488,727. 



7 _ 488,727/ _ 488,727 X 8 _ 



5. 



9,000 



= 434.424. 



260 



STRENGTH OP MATERIALS. 



But, — = ^^^ . ~ a" ^ '^^ ^^^' according to the conditions 

of the problem, b = — d. 

Therefore, - = i^^' = -^^^ = 434.424. 
' ^ 6 12 



^' = 5,213.088. 

d = nr- 1 

d = 8r. 3 



Ans. 



(713) The beam, with the moment and shear diagrams, 
is shown in Fig. 62. On the line, through the left reaction, 




Fig. 62. 



are laid off the loads in order. Thus, 6> i = 40 X 8 = 320 lb., 
is the uniform load between the left support and F^; 1-2 is 



STRENGTH OF MATERIALS. 2G1 

F^ = 2,000 lb. ; 2-3 =40 Xl2 = 480 lb., is the uniform load 
between F^ and F^; S-J,. = 2,000 X 1.3 = 2,600 lb., is F^, and 
4-5 = 40 X 10 = 400 lb., is the uniform load between F^ and 
the right support. The pole P is chosen and the rays drawn. 
Since the uniform load is very small compared with F^ and F^, 
it will be sufficiently accurate to consider the three portions 
of it concentrated at their respective centers of gravity 
Xj J, and ^. Drawing the equilibrium polygon parallel to 
the rays, we obtain the moment diagram a h b k c I d a. 
From P, drawing P in parallel to the closing line a d^ we 
obtain the reactions (9/;/ and in5^ equal, respectively, to 2,930 
and 2,870 lb. Ans. The shear axis in x^ and the shear dia- 
gram O r s t u V n in^ are drawn in the usual manner. The 
greatest shear is O m^ 2,930 lb. The shear line cuts the shear 
axis under F^. Hence, the maximum moment is under F^. By 
measurement, ^ris 64^, and/*jr is 5,000 lb. ; hence, the maxi- 
mum bending moment is 64 X 5,000 = 320,000 in. -lb. Ans. 

(714) From the table of Bending Moments, the great- 
est bending moment of such a beam is —5—, or, in this case, 



8 



w X 240' 
8 • 
By formula 117, 



^ ^£O<240^ _ 5/ _ 45^000 280 



8 fc 4 '" 12 ^ 2 

^, r 45,000 X 280 X 8 ^^ no iu • u ^ 

Therefore, w = — — ^ — — = 72.92 lb. per mch of 

240 X 240 X 4 X 6 ^ 

length = 72.92 X 12 = 875 lb. per foot of length. Ans. 

(715) From the table of Bending Moments, the maxi- 
mum bending moment is 



4 4 

From formula 117, 



= 24: IV. 

"A* 



r= ~(d*-d^*) = 56.U5; c =^d=^-i=3i; S=3S,000;/=^. 



263 STRENGTH OF MATERIALS. 

Tj »A ur 38,000 X 56.945 

«^"'=^' ^^ ^= G X 3.25 • 

^^ 38 000X5G.945 ^ .^^,^^ 
24 X X 3.2o 

(716) {a) From the table of Bending Moments, 

Af— !^- ^^ X 102' 
8 ~ 8 * 

From formula 117, 

^ ^X 192' __SJ 
'''- 8 "A' 

^. = 7,200;/=8; /= i^^^ ^ ^ ; ^ ^ ^^=. 5. 

o^v^.r. ^^ X 1^^' _ ^'^00 2,000 
men, ^ - ^ X i2:^^- 

_ 7,200 X 2,000 X 8 v_ 
^ ~ 8 X 12 X 5 X 192 X 192 ~ 
6.51 lb. per in. = 6.51 X 12 = 78.12 lb. per ft. Ans. 

(h\ T 1 ^^3 IPX 2^ 80 , 

{b)I=-bd =_^^=_ . = 1. 

w X 192' _ 7,200 80 

8 ~ 8 ^ 12 X 1' 

7,200 X 80 X 8 , ^ „ 

^ = T. — —r\ T7r\ ttt: = 1-3 lb. per m. 

8 X 12 X 192 X 192 ^ 

= 1.3 X 12 = 15.6 lb. per ft. Ans. 

(717) {a) From the table of Bending Moments, the 

5 W l^ 
deflection of a beam uniformly loaded is — — - „ , . In 

example 714, f^F^ 875 X 20 = 17,500 lb.; / = 240", E = 
25,000,000, and 7=280. 

Hence, deflection . = ggf ^/g^gooo^^^.go ='^' ^^' ^^^- 

1 Wl^ 
{b) From the table of Bending Moments, s = -~ ~^^' 



STRENGTH OF MATERIALS. 263 

In example 715, W= 4,624 lb. ; /= 96 in. ; ^=15,000,000, 

and 7^:56.945. 

„ 4,624x96^ 

Hence, s = -— , ^ '^ ^^,, ^Trrrr^ = -1 , nearly. Ans. 

' 48 X 15,000,000 X 56.945 ' ^ 



Wl 



(c) s = -~-^—.. In example 716 (a), 1^=78.12x16, 

/=192; ^5"= 1,500,000, and 7=-^^. 

1^ 

^ 5X78.12X16X192^ ,,.,„ . 

^^"^^> ^ ^ 384 X 1,500,000 X^n^ = -'^^- ^-^^^ " 

(718) Area of piston = \-7r d' = ^-7: x 14'. 

4 4 

W= pressure on piston = — - x 14' X 80. 

From the table of Bending Moments, the maximum 
bending moment for a cantilever uniformly loaded is 

w/' Wl i-Xl4'x80x4 5, 7 ,„ , iit^N 

—^r- = -TT- = ~ ?; = ~F -■ (See formulal 1 7.) 

2 2 2 / ^ 

5. _ 45,000 __ ^_jVfi^_l ^. 



Hence. 



i- X 14' X 80 X 4 4,500 :r^' 



2 32 



,3 14' X 80 X 4 X 32 ^^ ^^ 

or d^ = — -— — — ■ = 55.75. 

4 X 2 X 4,500 

^=='V^'55y75 = 3.82". Ans. 

(719) Substituting in formula 119, 5^= 90,000; A = 
6' X .7854; /=6; /= 14 X 12 = 168; ^=5,000; ^=^X 
6\ we obtain 

' 64 

5 

(720) For timber, 5, = 8,000 and /=8; hence, -^ = 

^ = 1,000. 



264 STRENGTH OF MATERIALS. 

Substituting in formula 108, 

P= A -^ = 1,000 A. 
. P y X 2,00 .. . - 

^ = i;ooo = 1,000 = ^^ '^- ^^^ ^^^^^^^^y ^^^^ «f 3 

j//^r/ column to support the given load. Since the column 
is quite long, assume it to be 6" square. Then, A = 36, 

Formula 119 gives 

/= 30 X 12 = 360, and^= 3,000. 
5, 14,000 A , 36 X 360^ 



' 36 X 360^ \ . Qon 1 

.^ + 3,000 X 108 j^'^^^^^^^^^^y- 



/ 36 

Since this value is much too large, the column must be 
made larger. Trying 9" square, ^ = 81, /= o46f. 

Then ^-^ - Hi22i/i 4_ 81 X 360 X 360 \ _ 
^^^""^ 7-~8^l^+ 3,000 X 546f j - ^'^^^' 

S 
This value of ~ is much nearer the required value, 1,000. 

Trying 10" square, A = 100, /= 1M22. = 833 J. 

5, 14,000 A , 100X300X360\ „.. , 

/ = ^10^ V + 3,000 X 833^ ) = ^'"'' ""^■■^y- 

vS 

Since this value of -- is less than 1,000, the column is a 

little too large ; hence, it is between 9 and 10 inches square. 

5 
Of will give 997.4 lb. as the value of ^; hence, the column 

should be 9|'' square. 

This problem may be more readily solved by formula 
120, which gives 

_ i / 7 X 2000 X 8 /7 X 2000 X 8 / 7 X 2000 X 8 12 x 360^ \ _ 
^~^ 3 X 800 "^ r 8000 ■ V 4 X 8000 "^ 3000 ) ~ 

>/?+ /Tr(3.5 + 518.4) = V 92.479 = 9.61" = H", nearly. 



STRENGTH OF MATERIALS. 265 

(721) Here 1^=21,000; /=10; 5, =: 150,000; g = 
6,250; /=7.5 X 12 = 90''. For using formula 121, we have 

.3183 Wf _ .3183 X 21,000 X 10 _ 

S^ ~ 150,000 --^^Sb. 

16/' 16X8100 



g 6250 

Therefore, 



= 20.7360. 



d= 1.4142 4^.4456 + 4/. 4450 (.4456 + 20. 7360) = 
I.41424/.4456 + 3.0722 = 2.65", or say ^". 

(722) For this case, A = 3.1416 sq. in. ; /= 4 X 12 = 
48''; 5, = 55,000; /= 10; /= .7854; ^= 20,250. 

Substituting these values in formula 11 9, 

j^_ S\A _ 55,000 X 3.1416 __ 

~ f(l I ^n~io/l I 3.1416X48- \- 
•^l "^^// ^T~^ 20,250 X. 7854; 
5,500 X 3.1416 
1.4551 
Steam pressure = 60 lb. per sq. in. 

rj^u ^ • . ^o-A ^. ^ 5'500 X 3.1416 

Then, area of piston = .78o4 d^ ~ -— — \ — — . 

' ^ GO 1.4551 X 60 

Hence, d'"" = ^,^^V , , ' , — = 252, nearly, 

' .7854 X 1.4o51 X 60 ' ^' 

and <^= 4/252 = 15|-", nearly. Ans. 

(723) (a) The strength of a beam varies directly as the 
width and square of the depth and inversely as the length. 

Hence, the ratio between the loads is 

6 X 8\ 4 X 12^^ -, r> 1 ^ 11 A 
— ^— : ^^ = 16 ; 15, or l^V Ans. 

yU) The deflections vary directly as the cube of the 
lengths, and inversely as the breadths and cubes of the 
depths. 

Hence, the ratio between the deflections is 

10^ 16= 



6X8' 4 X 12= 
D. 0. l!t.-32 



= .549. Ans. 



2«6 STRENGTH OF MATERIALS. 

(724) Substituting the value of c^^ from Table 31, in 
formula 123, we obtain 

(a) ^=^,/^ = 4.92f^ = 3.739". Ans. 



= .,/5 = 4.92l/^ = 4.65^ Ans. 



(b) d-.,, ^--^.^~r ^00 

(725) Using formula 123, 



//= 4.7^1/^^^;^ = 14.06". 



'4,000 



50 



Since this result is greater than 13.6", formula 124 must 
be used, in which 



^=^^y^=3.3f?2p = 14.22". Ans. 

(726) From formula 123, 

d=c^\^^, or H=^^. c^ = 4.11. (Table 31.) 

Hence, H = \ , , , = 71. 775 H. P. Ans. 
' 4.11 

(727) Using formula 126, 

i7=^ W^^'^^ = .0212xlOo(^^?^i^^'W717.7 H.P. Ans. 

(.0212 is the value of q^ from Table 32.) 

(728) (^) Using formula 1 27, 

(P= 100 C = 100 X 8' = 6,400 lb. Ans. 

{d) Using formula 128, 

i^-600d:^ or^^--^- = M2^ = 10 

6-= 4/10 = 3.162". 

^=i(f= 1.054". Ans. 
o 

(c) Using formula 1 29, 

C = VT3^ = 3. 651". Ans. 



STRENGTH OF MATERIALS. 

(729) (a) Using formula 130, 

P= 12,000^' = 12,000 X (~) = 0,187-5 lb. Ans 
(d) Formula 131 gives P= 18,000^'. 
Therefore, d 



267 



4/=4==/^ 



18,000 



,000 
18,000 



= .667^ Ans. 



(730) The deflection is by formula 118, 

iv/^ 1 W/'' 
s = a -y^-r = -^T^ ■ r- r i the coefficient being found from the 
ii / 192 A I 

table of Bending Moments. 

Transposing, IV =^^^, ; /=120; ^ = 30,000,000; / = 



/= 



.7854; s = ^. 



Then, W= ''' X ''f''Z ^ ■'''' = 327.25 lb. Ans. 

' 8 X 120 



(731) {a) The maximum bending moment is, accord- 

mg to the table of Bendmg Moments, — — = = 

90,000 inch-pounds. 

By formula 117, 



i^= 90,000 = -^-. 



Tzd' 



6. = 120,000;/= 10. - = ^ = — . 

120,000 TuP 



Hence, 90,000 = 



10 



32 ' 



or d 



^f^ 



),000 X 10 X 32 



120,000 X 3.1416 
(i) Using formula 123, 



= 4.244'' = 4^", nearly. 



d=c^ |/J := 4. 7 i/| -. H'\ nearly. Ans. 



268 



STRENGTH OF MATERIALS. 



(732) {ci) The graphic solution is shown in Fig. 63. 
On the vertical through the support 6^ i is laid off equal to 
the uniform load between the support and A, : /-? is laid off 




Scale of forces 1^800 Ih. 
Scale of distance 1—4^ 

Fig. 63. 

to represent F^. 2-3 represents to the same scale the re- 
mainder of the uniform load, and 8 i;i represents F^. The 
pole P is chosen and the rays drawn. The polygon a h c e fh 
is then drawn, the sides being parallel, respectively, to the 
corresponding rays. If the uniform load between F^ and F^ 
be considered as concentrated at its center of gravity, the 
polygon will follow the broken line c c f. It will be better 
in this case to divide the uniform load into several parts, 
2-Ji.^ 4-5, 5-6, etc., thus obtaining the line of the polygon 



STRENGTH OF MATERIALS. 269 

c d f. To draw the shear diagram, project the point 1 
across the vertical through i^,, and draw O s. Next project 
the point 2 across to /, and 3 across to ?/, and draw / ?/. 
O s t ?i 71 in is the shear diagram. The maximum moment 
is seen to be at the support, and is equal to a Jiy^ P in. To 
the scale of distances, a /i =z oB-S in., while P in = 1/ ~ 
1,400 lb. to the scale of forces. Hence, the maximum bend- 
ing moment is 58.8 X 1,400 = 82,320 in. -lb. Ans. 
(^) From formula 117, 

6" / 
M = -^ - = 82,320. 5, = 12,500;/= 8. 

Therefore, - = ^^;f ?^^ ^ = 52.68. 
c 12,^00 

But, - = ^\ J = -— -. and ^/ = 21 ^, or <^ = -^ . 

Hence, - = --- = -^ = 52.68. d^ = 52.68 X 15 = 790.2. 
<r D 15 

2d 



d = |/790.2 = 9. 245". b — -—■ = 3. 7^ nearly. Ans. 

o 

(733) Referring to the table of Moments of Inertia, 
{bd-^ - h^ d^y -^bdb^ d^d- <)' _ 
\2{bd-bJ^) 
[8 X 10^ - 6 X {8\YY - 4 X 8 X 10 X 6 X Sj (10 - Sj )' __ 
12 (8 X 10 - 6 X 8^-) 
280.466. 

d b,dj d-d, \_ 
2"^ 2 \bd-b,dj~ 
10 6x8i / 10-8i \ 0310 
2^ + ~T"^l8 X 10- 6 X 8i; - ^•^^^- 
{a) From the table of Bending Moments, the maximum 

u ^- • '^Vl 

bendmof moment is — - — . 

4 

S^=z 120,000; /=7; /=35X 12 = 420 in. 

Using formula 117* M= —— — -J—, or 
=* ' 4 jc 

4SJ^ 4 X 120,000 X 280.466 ^ ' 

//c 420 X 7 X 6.319 



2?0 



STRENGTH OF MATERIALS. 



(d) In this case /=5, and the maximum bending 
Hence, from formula 11 7, 



. w I 
moment is -— - 



M = 



w /' 



sr 



sr 



= — - — or zu = — 



rr.u f Tjr ; SSr SX 120,000 X 280.466 

Therefore, lV=wl= , * = ' = 

Ifc 420 X 5 X 6.319 

20,290 lb. Ans. 

(734) {a) According to formula 115, 

/= A r\ or r= f — = f || = 4/8 = 1.732. Ans. 




Scale of forces 1=960 lb. 
Scale of distance 1=8^ 

Fig. 64. 

('') From the table of Moments of Inertia, I = ~bd* = 

72 ; ^ = ^ ^ == 24. Dividing, -^^f^- = ^, or ^-cP ^ 3. d\= 

b a 24 12 

36: ^=6" and Z' = 4". Ans. 



STRENGTH OF MATERIALS. 271 



(c) As above, r = \/4 = a/ ~ = V^= i Ans, 




A y Tzd' ^ 16 4' 

^ d 

(735) Using formula 112,/<^= 4/6", we have /=^-^. 

Using a factor of safety of G, 

^^=^'°'-^ = T5-= 4X20,000 =-°'^ ^"^- 

(736) The graphic solution is shown in Fig. G4. The 
uniform load is divided into 14 equal parts, and lines drawn 
through the center of gravity of each part. These loads 
are laid off on the line through the left reaction, the pole P 
chosen, and the rays drawn. The polygon be d e fa is then 
drawn in the usual manner. The shear diagram is drawn 
as shown. The maximum shear is either t 7 or r v — 540 lb. 
The maximum moment is shown by the polygon to be at 
f c vertically above the point ?/, where the shear line crosses 
the shear axis. The pole distance P 7 is 1,440 lb. to the 
scale of forces, and the intercept/^ is 14 inches to the scale 
of distances. Hence, the bending moment is 20,160 in. -lb. 

(737) From formula 117, 

5 / 

M= y- = 20,160. 5, = 9,000; /= 8. 

Thel ^- ^Q.1GQX8 _ 
^^^""^c - 9,000 -^^•^^• 

/ "^ d d^ 1 
But, — = ^^, — r~ = —bd'' for a rectangle. 
c ^d b 

Hence, )rbd'' = 17.92, or bd'=: 107.52. 

Any number of beams will fulfil this condition. 

Assuming d=6\ d = — ^ — = 3", nearly. 

oh 

Assuming d= 5", b = ^^^^ = 4.3". 

(738) Using the factor of safety 10, in formula 114, 
_ 9,600,000 ^ ^ 960.000 X .2-' ^ ^^ ^„^ 

' ^ 10 /d 108 X 2.5 



272 



STRENGTH OF MATERIALS. 



- (739) Using formula 130, 

/^= 12,000 ^^ or ^= j/-^- = '/?XM22 = .913^ Ans. 

^ 12,000 ^ 12,000 '^^^ ' ^ ^' 

(740) The radius r of the gear-wheel is 24". Using 
formula 123, d= c^Pr = .297 ^'S50 X 24 = 2.84". Ans. 

(741) Area of cylinder = .7854 X 12' = 113.1 sq. in. 

Total pressure on the head = 113.1 X 90 = 10,179 lb. 

10 179 
Pressures on each bolt = — '— — = 1,017.9 lb. 

Using formula 108, 

P 1 017 Q 
P= AS, or A =^= -^^^ = .5089 sq. in., area of bolt. 



5 2,000 
Diameter of bolt = y 



5089 

.7854 



= . 8", nearly. Ans. 



(742) (a) The graphic solution is clearly shown in Fig. 
65. On -the vertical through P^, the equal loads P^ and P^ 




Scale of forces 1=2000 lb. 
Scale of distance 1=6^ 

Fig. 65. 
are laid orf to scale, O 1 representing P^ and 1-2 representing 



I 



STRENGTH OF MATERIALS. 273 

F^. Choose the pole P, and draw the rays P O^ Pi, P 2. 
Draw a b between the left support and F ^ parallel \,o P O \ 
b c between F ^ and F ^ parallel to Pi, and c d parallel to 
P2, between F ^ and the right support. Through P draw a 
line parallel to the closing line ad. 1-= 1-2 \ hence, the 
reactions of the supports are equal, and are each equal to 1 
ton. The shear between the left reaction and F^ is nega- 
tive, and equal to /^^ = 1 ton. Between the left and the 
right support it is 0, and between the latter and F^ it is posi- 
tive and equal to 1 ton. The bending moment is constant 
and a maximum between the supports. To the scale of 
forces /^i = 2 tons = 4,000 lb., and to the scale of distances 
<3: y= 30 in. Hence, the maximum bending is 4,000 X 30 
= 120,000 in. -lb. Ans. 

(U) Using formula 117, 

S" / 
M- -F - = 120,000. S^ = 38,000; /= 6. 
/ ^ 

^, / 120,000X6 360 ,^ , 

^^"^' 7 = -^8;000- = -19- = ^'' ^"^^^y- 

^ d' 



_ ^ / 64 Ttd'' 



c ^d_ 32 * 
2 

Hence, _ = 19, or ^ = -33^. 



^//32X19 



(743) Since the deflections are directly as the cubes of 
the lengths, and inversely as the breadths and the cubes 
of the depths, their ratio in this case is 

18' 12' 27 9 ,^ 

or -- : — = 12. 



2X6' 3X8" 2 

That is, the first beam deflects 12 times as much as the 
second. Hence, the required deflection of the second beam 
is 3 -^ 12 = .025". Ans. 



274 STRENGTH OF MATERIALS. 

(744) The key has a shearing stress exerted on two 
sections; hence, each section must withstand a stress of 

— == 10,000 pounds. 

Z 

Using formula 108, with a factor of safety of 10, 

r, AS^ . 10 P 10X10,000 ^ 

^-^-,or^^^:^ 50,000 -^^q-^^' 

Let d = width of key ; 
/ = thickness. 

Then, ^ / = ^4 = 2 sq. in. But, from the conditions of the 
problem, 



Hence, ^/ = i^'' = 2; d' = S; ^ = 2.828^ 

- 4: 



^ Ans. 



(745) From formula 118, the deflection s = a ^ ^ , 
and, from the table of Bending Moments, the coefficient a for 

the beam in question is ■— . 

^ 48 

lV=dO tons = 60,000 lb. ; /= 54 inches; £ = 30,000,000; 



/ = 



G4* 



^ IX 60,000 X 54' ^^^, . 

Hence, s = '- =.0064 m. 

48 X 30,000,000 X g^ Ans. 

(746) (a) The circumference of a 7- strand rope is 3 
times the diameter; hence, 6' = 1^ X 3 = 3f". 

Using formula 129, 7^=1,000 C = 1,000 X (S^Y = 

14,062.5 lb. Ans. 

(b) Using formula 127, 



STRENGTH OF MATERIALS. 



275 



(T47) Using formula 113, 



/ = 



St 120,000 



r+ t 



= 12,000 lb. Ans. 



+ 6 



(748) The construction of the diagram of bending 









F, 



F^ 



F4 



1 



-20^ — 'l^ 20^- 



m2'' 



f/"' 



Scale of forces 1=1600 lb. 
Scale of distance 1=32' 

Fig. 66. 




276 STRENGTH OF MATERIALS. 

moments and shear diagram is clearly shown in Fig. GG. It 
is so nearly like that of Fig. 58 that a detailed description 
is unnecessary. It will be noticed that between k and k' 
the shear is zero, and that since the reactions are equal the 
shear at either support = -|- of the load = 2,400 lb. The 
greatest intercept is r r' = <^<3^' = 30 ft. The pole distance 
77= 2,400 lb. Hence, the bending moment = 2,400 X 30 = 
72,000 ft. -lb. =: 72,000 X 12 = 864,000 in. -lb. 



APPLIED MECHANICS. 



(QUESTIONS 749-808.) 



(749) See Fig. 341 in Art. 1434. The construction 
is shown in Fig. 67. 




(750) The arms should be in the proportion 3 : 4. See 
Fig. 345 of Art. 1438. 

(751) In Fig. 68, let A B and C Dh^ the two center lines 
of motion, making an angle of 10° with each other. Draw 
a b parallel to A B, and at a distance of 3" from it ; also, c d 
parallel to C D, and at a distance of 4" from it. O^ the point 
of intersection, will be the center of the lever, and O E and 
O H will be the center lines of the arms when in mid-position. 

(752) The lengths of the lever arms will be as 3 X 12: 
3.5::36:3.5. Either the motion shown in Fig. 343 or one 
of those shown in Fig. 344 may be used. 

For notice of copyright, see page immediately following the title page. 



i 



278 
(753) 





APPLIED MECHANICS. 

Circumference of pinion = 12 X 3.1416 = 37.699". 

Distance each moves = 
6.5 




37.699 X 



= .68". Ans. 



360 

(754) The construction 
is shown in Fig. 69. This, 
in connection with the prin- 
ciples explained in Arts. 
1432 to 1438, should 
enable the student to lay 
out the motion. 

(755) See Art. 1441. 

(756) The forward 
stroke, in this case, is the 
stroke from left to right, 

while the crank-pin moves through the lower part of its 
circle. Cross-head A leads during the forward stroke, and 

Half Cir. 1.571'^ 



Fig. 68. 




Fig. 69 



cross-head B during the return stroke. The construction 
to be used in determininsf this is shown in Fig. 349. 



APPLIED MECHANICS. 



279 



(757) The effect is the same as though one toggle-joint 
were used instead of two. First, find the pressure exerted 
by the screw. The distance that the power moves in one 
turn of the lever = 20 X 3.1416 = 62.832"; distance that the 

r>i\ Q Q O 

nuts move = ^"; velocity ratio = — ^— = 251.328; force 

exerted =: 75 X 251.328 = 18,849.6 lb. 

When distance B = l(j", the distance corresponding to /i. 



in Fig. 352, = 



16 - 12 

2 



= 2'^ The distance corresponding to 

Whence, 



O r, therefore, equals \fW- V = |/320 = 17 
applying formula 135, 



P = 



18,849.6 X 17.888 
2X2 



= 84,295.4 lb. Ans. 



(758) In the diagram in Fig. 70, HE, C E, E R, and 




Fig. 70. 



OR are center lines of the parts which have the same letters 
in Fig. 22, question No. 758. 



280 



APPLIED MECHANICS. 



Draw O d perpendicular to E R produced ; then by the prin- 
ciple of moments (see Art. 1447"), the thrust in the direc- 



tion oi E R = 



'lY.Or 2X1.5 



= 6.82 lb. Erect the per- 



Od .44 

pendicular C e^ and draw C f perpendicular to H E and C g 
perpendicular to H C. Then, the pull on the pin at H will 

.6.82x6^^ 6.82X1.28 ,, ^q i^ a tj • .i 

equal y^-j = — = 14.08 lb. Ans. Horizontal 

6.82 X Ce 



Cf 
thrust of cross-head = 



6.82 X 1.28 
.63 



13=86 lb. 
Ans. 



Cg 

(759) See Art. 1457. 

(760) The construction is shown in Fig. 71. The 

various points are lettered in 
the same manner as in Fig. 358, 
and the description in Art. 
1452 will apply to Fig. 71 in 
all respects, except that the 
circle must be divided into five 
equal parts instead of three, 
and the arc deb must contain 
three of these parts in order to 
have the ratio 3 : 2. 




Scale f=l ft. 



(761) {d) Fig. 72 (^) shows 
a diagram giving the propor- 
tions of the various parts; the 
arc ^ ^^is one-half as long as 
the arc a c g. See Arts. 1454 
and 1455. 

[b) 360 -^ 20 = 18 ; hence, the 
crank-pin circle d c b e must be 
divided into 18 equal parts. 
The motion is shown in Fig. 72 (b). See Art. 1455. 

(762) See Art. 1459. {a) Greatest rate = 50 X 

— = — -=i — 56.63 rev. Least rate = 50 X .88295 = 

cos 28° .88295 

44.15 rev. 

{b) Connecting as described would double the variation 



Fig. 71. 



APPLIED MECHANICS. 



281 




'At 



\ 1 



-6 



u 



30 



D. 0. III.— 33 



8 

Fig. 73. 



282 



APPLIED MECHANICS. 



(763) In Fig. 73, which is lettered the same as Fig. 364, 
draw the given center lines and locate points n and A from 
the data of the problem. Lay off in a ^ \ stroke ; also 
n d ^ \ stroke. Connect A and a and draw a B perpendicu- 
lar to A a. Through (9, draw^ O C\ connect C and b, and 
through b draw b D perpendicular to C b. Measure the 
lengths C D and A B. 

(764) The construction of the cam is shown in Fig. 74. 




(765) The construction of the cam is shown in Fig. 75. 
It is similar in principle to the one shown in Fig. 366. 

(766) The shape of the development of the groove is 
shown in Fig. 76. The method of laying it out is also 
indicated. 



APPLIED MECHANICS. 



283 



(767) This cam is shown in Fig. 77, and is like the one 
shown in Fig. 369 in all respects except that the dimensions 
are different. The description given in Art. 1474 will 
apply to this case. 




Fig. 75. 



(768) See Art. 1478. 
min. Ans. 



52 X 320 

48 



= 346f rev. per 




Fig. 76. 



(769) 



500 X 10 
210 



= 23.8" = 23f . Ans. 



284 



APPLIED MECHANICS. 



(770) Substituting in formula 137, 189 X A X 20 = 
1,800 X 6 X 3.5, or 

_ 1,800X 6X3.5 _ 
• ^^- 189 X 20 -^^' ^''^' 




(771) See Art. 1481, 900 -^ 150= 6; 2 x 3 = 6, and 
4 X 1^ = 6. The ratios would then be 2 : 1 and 3 : 1; also 
4 : 1 and 1|- : 1. Other ratios could be used. 

(772) See Art. 1483. (a) Effective pull in foot- 
5 X 33,000 



pounds 



200 X 3.1416 X 3 



87.54 lb., nearly. Ans. 



/ix T^- . ' • u 5 X 33,000 X 12 no ' u 

(3) Diameter m mches = ^^-^^^^^^^^^^^ = 63 mches. 

Ans. 
(773) By laying out the two pulleys, the arc of contact is 
found to be about 150°. The allowable effective pull per inch* 
in width, taken from the table is, therefore, 33.8 lb. Ans. 



APPLIED MECHANICS. 285 

The effective pull of the belt = ^ X 33,000 X 12 ^ 
^ 4X3.1416X1,500 

ib. 126.05 -^ 33.8 = 3.7 inches. A 4-inch belt would be 
used. Ans. 

(774) The speed of the belt in feet per minute ~ 
14 X 3. 1416 X 90. Substituting in formula 142, 

„. 630 X 250 

^= Ti ^TTTTT^ 7^ — ^^ , nearly. 

14 X 3.1416 X 90 ' ^ 

The wheel should be 41 inches wide, Ans. 

(775) Speed of belt = 2 X 205 X 3.1416. Substituting 

• f 1 -ioo zT 6X2X205X3.1416 ^ ^ tt -d 
m formula 138, H= — — = 8.6 H. P., 

t/UU 

nearly. Ans. 

(776) Calculating first for a double belt, we find that 

„ 6X2X205X3.1416 ,^ ^ , 

H = — = 12. 3, nearly. 

Next, calculating for a single belt with 30-inch pulley, we 

find that 

^ 6X2.5X205X3.1416 ,^ ^ . 

H= — -=10.7, nearly. 

The double belt would, therefore, be the more effectual 
remedy. Ans. 

(777) {a) Substituting in formula 147, 



18 = ^r ^; or d= 18 ^ 4/5.7143 = 7.53 inches. Ans. 



(d) Substituting in formula 146, N= fTOxiOO = 167 j 
rev. per min. Ans. 

(778) (a) Substituting in formula 144, 

Ratio, therefore, =1.78 : 1. Ans. 
(b) Substituting in formula 146, 

225' 

225 = ^400 X n^ , or n^ = — - = 126.56 rev. per min. Ans 



286 



APPLIED MECHANICS. 



(779) First calculate the middle diameter, using formula 
148. 




Fig. 78. 

,^ 15 + 4.5 , .08(15-4.5)'' ^ ^^ , ,, ,^,^. , 
M~ 1 ^^ — -^ ~=^ 9.75 + .44 = 10.19 mches. 

One of the pulleys is shown in Fig. 78. 

WlUtHM^ fW%wW^ (^^^) See Fig. 79. 
^ * An 8-inch belt should 

have a double row of 

holes, five in the row 

nearest the end. Space 

the holes equally. 






{ 



VVvHVUMWWil 




^ 



/v\#w%vWw 



Fig, 79, 



(781) The ar- 
rangement of the pul- 
leys is shown in Fig. 
80. The belt leads 
from point ^, on pul- 
ley Ay into the plane 



APPLIED MECHANICS. 



287 



of pulley B ; and from point b, on pulley j5, into the plane 
of pulley A. 

(782) Fig. 81 shows how the two shafts may be connected. 
The arrangement is entirely similar 
to that shown in Fig. 392. 

/«rco\ 60x30x60x30 , 

(^^^) - 15 X 15 X 10 = ^^"^^^^ 
of revolutions of last shaft = 1,440 
rev. per min. Ans. 

The crossed belt and gears each 
reverse the direction of rotation; 
the open belt does not change the 
direction. Hence, the first and last 
shafts turn in the same direction. 

(784) Gears B and C^ being 
idlers, do not affect the number of 
rotations made by D. Hence, D 

makes — = 1 turn for every turn of 

A. As the number of axes is even, 
the first and last turn in opposite 
directions. fk^ go 

(785) Referring to Fig. 398, and applying formula 

137, we have N X 
^^f, X < = ?2 X /, X 
yjj, N and n being 
the number of revo- 
lutions of the spin- 
dle and lead screw, 
respectively. Solving 
for ^„ 

''^- Nxd^ ' 

Now, to cut 4 

threads per inch, the 

spindle must turn 4 

Hence, sub- 





Fig. 81. 

times while the lead screw turns 8 times. 



288 APPLIED MECHANICS. 

stituting in the formula, we have n = S,/^ = 2, JV= 4, <^i = 1, 

Since Icy, remains constant for this lathe, we may 



JV • 



hereafter write the formula as d^ = 

For 5 threads, d^ = — ^ = -^/a- 

For 6 threads, < = i^ = |-/,. 

For 7 threads, d^ = —z~ = -wf^- 

For 8 threads, d^ = — ^ = 2 /, , etc. 

o 

To make out the table, we should aim to use the same 
gears as many times as possible. Start with 96 teeth for 

9(3 
the gear on stud. Then,/^ = — = 24 teeth. For 5 threads, 

v q 

/^ = — X 96 = 30, when d^ = 96; for 6 threads, /, = - X 
lb o 

7 
96 = 36, when /^, = 96 ; for 7 threads, /, = -^ x 96 = 42 ; for 

8 threads, /^ = - x 96 = 48. The other gears are found 
in the same manner, 

(786) The gear makes |- a turn in passing the center at 
each end of the rack, two turns on top of the rack, and two 
turns while traveling over the under side. 

(^?) The circumference of the gear — 10 X 3.1416 = 
31.416". Distance that the rack travels uniformly = 2 X 
31.416 = 62.832"= 5 ft. 2.832". Ans. 

(b) The motion is harmonic as the gear passes over the 
ends, and the horizontal distance traveled by the rack at 
eacn end = the radius of the gear; distance traveled at both 
ends = 2 X radius, or 10 inches. Hence, 5 ft. 2.832"-|- 10" 
= 6 ft. .832", the total travel. Ans. 



APPLIED MECHANICS. 28S 

(787) When driving through the worm and worm- 
wheel, 40 turns of the worm produce one turn of the driven 
shaft T. When driving through the gearing, 40 turns of 

40 X 80 
the shaft vS produce — — = IGO turns of T, Hence, the 

ratio of the " quick return " is IGO : 1. 

(788) Circumference of gear Z>,==3.5X 3.1410 = 10. 9956 = 

11", nearly. Hence, for every foot travel of the table, D^ 

12 
makes — = ljJy revolutions. To find the number of turns 

made by the pulley for each foot passed over by the table, 

we have, therefore, A^ X 3. 5 = ^ + 26, or A^ = ,, ,^ = 

11 11 X 0.5 

8.103 -[-• This, multiplied by the circumference of the 

,. 8.103X30X3.1416 .. « u .i. .• • 

pulley, = -— = 63.6. Hence, the ratio is 

63.6 : 1. Ans. 

(789) (a) D F A , 



Wheels locked +10+10 +10 

Arm stationary — 10 — 10 X -^ 



+10-10X^^ +10 



Number turns of /^= 10 '-^-^ = —.101 turn. 

The wheel /, being an idler, it is not considered. 
(^) D F A 



Wheels locked +10+10 +10 

-.0 -.ox!52 



+10-I0xi^y +10 



Number turns of /^= 10 - ^^^ = + .099. Ans. 



290 APPLIED MECHANICS. 



(790) {a) D 


F 


A 


Wheels locked -f 4 


+ 4 


+ 4 


Arm stationary — 4 


-h« 





Arm stationary + 1 


+ ii 





+ 1 


+ i 


+ 4 


/■=+4-5 + li=+i. 


Ans. 




(b) D 


F 


A 


Wheels locked -\- 5 


+ 5 


+ 5 


Arm stationary — 5 


-ix» 





Arm stationary — 1 


-H 






-1 -^ +5 

F= + 5 - 6i - li = - 2^. Ans. 

(791) D F, D^ F^ £ 

Train locked +6 +6 +6 -|-6 

40 
Arm fixed - 6 - 60 + — X 60 



-54+126 +6 

20 X 12 X 10 



(792) 20 turns of shaft 5 will produce 
8 turns of D, L. H. Hence, 



3 X 100 



E 


H,K 


D 


-8 


-8 


-8 


+ 8 


-8 





+ 20 


-20 






Wheels locked 
Arm fixed 
Arm fixed 

+ 20 - 36 - 8 
Pulley K makes — 36 revolutions. Ans. 

(793) See Arts. 1550 and 1558. 

(794) {a) See Art. 1552. 

(b) Substituting in formula 1 50, 

,. 3.1416 3.1416X2 ..^^ . , . 

6= — -; — = 7^ = .698 mch. Ans. 

4i 9 

{c\ Substituting in formula 149, P= t^zttttt = ^f- Ans. 
\ / & ' 1.1424 



APPLIED MECHANICS. 



291 



(795) (a) From formula 151, 

i>'= -p =-r- = 10 mches. Ans. 

(^) Substituting in formula 1 52, 

0/)=. J = 10. 667 inches. Ans. 
o 

(c) Addendum = -=.3333"; working depth = 2 X .3333 - 
.667 inch. 

(d) Clearance = .3333 X i- = .042^^; whole depth = .667+ 
,042 = .709". Ans. 




FtG. 82. 



(796) See Fig. 414. 

(797) (a) Number of teeth = 48 X 10 = 480. Ans 

{b) Number of teeth = (14— 2xi)x6 = 82. Ans. 



292 



APPLIED MECHANICS. 



(798) From formula 1 53, i^ = ^fui^ = ^^- ^^^ 

(799) Substituting in formula 154, 



a = 



2AV 2X15.5X2 62 



V-i-2' 2 + 3 5 

Substituting in formula 155, Z> = 
L8| inches. Ans. 



= 12| inches. Ans. 

2 X 15.5 X 3_ 93 
5 



3 + 2 




Fig. 83. 

(800) (a) See Art. 1560. 
(^) See Art. 1563. 

(801) See Art. 1573. 



APPLIED MECHANICS. 



293 



(802) Diameter of internal gear =: 30 -^ 3 = 10 inches. 

Diameter of pinion = 21 -^ 3 = 7 inches. 

10 — 7 
Diameter of describing circle = — - — = 1^ inches. 

See Art. 1 565. 

(803) To determine whether two involute gears will in- 
terfere, use the construction shown in Fig. 425. In Fig. 82, 
it will be observed that the addendum circles of both gears 
cut the line of action iV JV^ at points e and d, outside of 
points E and D. Hence, both gears will interfere. 

(804) The solution to this example is shown in Fig. 83. 

(805) Since the worm is triple-threaded, it will drive 
the wheel three teeth at every revolution; in 100 revolu- 
tions, it will drive it 100 X 3 = 300 teeth, or — — =7.5 revo- 

40 

lutions. Ans. 

(806) (a) The wheel must have the same circular pitch 
as the worm, or f'', and, as the worm is single-threaded, the 
number of teeth in the wheel is the same as the ratio of 



X 40 



= 9.55". From Table 



speeds, or 40. Hence, D= n ^A^a 

34, the addendum = .3 X f =.23^; outside diameter = 9.55 + 
(2X .23) = 10.01, or, say 10'; whole depth, .23 + (.35 X f ) = 
.23-J-.26 = .49^ 




(b) The section of the worm is shown in Fig. 84. 



294 APPLIED MECHANICS. 

(807) See Art. 1581. Tangent of angle of thread = 

circum. of cylinder 2x3.1416 ^ „ o-..-.^ -.n o.r^/» 
— ■ = 2 X 3 X 3.1416 = 18.8496. 



pitch 



The angle, therefore, equals 86° 58', nearly. Ans. 

(808) For every tooth moved, the screw would make 
y^2 of a turn. For every turn of the screw, the "feed" 

would be ji^ = I inch. Hence, -^^ x ^ = ^|^ = g^g inch 

= the required "feed." Ans. 



MACHINE DESIGN. 

(QUESTIONS 959-1023.) 



(959) See Art. 1910. 

(960) See Art. 1911. 

(961) (a) See Art. 1915. 
(b) See Art. 1912. 

(962) See Arts 1924, 1925, 1927, and 1928. 

(963) From Table 43 the number of threads per inch 
corresponding to the diameters given are 13, 9, 7, and 4-^, 
respectively. Applying formula 21 1 : 

^=^-.— "= 1-^"= .4". Ans. 

< = ^-— "= ^-i^"= .73r. Ans. 

* n 8 d 

d=d-—'=:l\^^''= 1.064". Ans. 

< = ^~ — '^ = 2 - \^"= 1.7ir. Ans. 

* n 4J 

(964) {a) Apply formula 208. 



For i" diameter, / = .24 \/ .25 + .625 - .175" = .0495^. 
For f diameter, / = .24 4/. 625 + .625 - .175'' = .0933^ 
For V diameter, / = .24 \/ 1 + .625 - .175" = .131". 
For \Y diameter,/ = .24 \/ 1.5+ .625 - .175" = .175". 
Now, applying formula 21 0, 

n = — : — = 20.2, say 20 threads per inch. Ans. 
.0495 

»=: = 10.7, say 11 threads per inch. Ans. 

•V/t/OO 

For notice of copyright, see page immediately following the title page. 



320 MACHINE DESIGN. 

w= — —r = 7. G, say 8 threads per inch. Ans. 

. Xo-L 

n^ -— — —• 5.7, say 6 threads per inch. Ans. 
.175 

(b) By iormula 210, ^^ = t; or, p — -, Hence, 

1 

p — —" — pitch for ^' screw. Ans. 

/ = — " = pitch for I" screw. Ans. 
/ = Q " = pitch for \" screw. Ans. 

o 

/ = t:" — pitch for 1^* screw. Ans. 
i965) (a) See Art. 1931. 
(^) \-^^-\' Ans. 

(966) See Art. 1930. 

(967) From Table 25, the ultimate tensile strength of 
wrought iron is 55,000 lb. per sq. in. From Table 28, the 
factor of safety for a varying stress is 6. Hence, the safe 

stress per square inch is ^, and 11,800 ^ ^^ = 

— ' ^^^ — = 1.2873 sq. in. = area at bottom of thread. 
00,000 

From Table 43, the nearest area in the last column is 1.2928 

sq. in., corresponding to a nominal diameter of 1^". Hence, 

the bolt should have a diameter of 1^". Ans. 

(968) Using formula 217, 

^^-6;ooo = ^-''^^^-^"- 

From Table 43, the nearest diameter corresponding to this 
area is \\" . Ans. 

The reason that this diameter should be used instead of 
that calculated in the last example is that the bolt is greatly 
weakened by the sharp corners left by the thread cutting tool, 



MACHINE DESIGN. 



321 



and the strength is not equal to that of a round bar whose 
diameter is the same as the diameter at the bottom of the 
thread. 

(969) Using formula 218, 

d - .0228 4/12,000 = 2^". Ans. 

^ = 2.5 X I = 2". Ans. 
o 



Using formula 219, 

12,000 



n. 



= 10 threads. Ans. 



300 X 2' 

(970) The dimensions of both head and nut are the 
same. Hence, apply formulas 220-226. 
Height of head or nut = 



^' ~ 8 " 16 ~ 16 • 



Ans. 
Diameter of nut or head across flats = 

Ans. 



17 1 
^ ~ -^2 ^ 8 + 16 ■ 



8 ' 



Diameter of nut or head across corners = 

n^ = 1.73 X |-' + .07^ = 1.58^ Ans. 



Diameter of washer = 



1 OK 

i^3 = 1^ X 1.58'' = 1.78^ say l|^". 



Ans. 



Thickness of washer = 



= .15 x|'=.13% say ^^ Ans. 

o o 



Fig. 605 shows the manner of representing the bolt, ex- 
cept that a hexagonal head is to be shown instead of a 
square one, and the letters are to be replaced by the sizes 
calculated. 

(971) In the last example, D was found to be If", The 
other dimensions are readily obtained from Fig. 606. The 
student should give the handle of the wrench a slight taper 
to improve the looks. 

D. 0. 111.-34 



322 MACHINE DESIGN. 

(972) From Table 28, Art 1362, the factor of safety 

for a varying stress is 6 for wrought iron. Hence, area of 

55 000 
cross-section of eye bolt = 2,500 ~, ^— — = .273 sq. in. The 



= d=/--^'' - '^ 



corresponding diameter = d = y ' = . 59", say — ". Ans. 

.7854: o 

Applying formula 227, 

d^ = .Sx~=~\ Ans. 

1 1 /5\' 

Taking a b equal to ^^', ab= - X (5-) = .195 sq. in. 

/C yO \8 / 

If we assume that ^ = — ^, ab •= —b'^y and b = 4/. 195x2 = 

.624^say|^ Ans. 
o 

15 5 
Whence, a = ~x x- = zrx". Ans. 

/& o ib 

(973) See Art. 1944. 

(974) See Fig. 623. Take height of thin nut as 
one-half that of outer or standard size nut. 

(975) Both devices increase the friction between the 
threads of the screw and the threads in the nut; in this 
respect the principle is the same in both devices. 

(976) Taking d as IW the various dimensions are 
readily obtained from Fig. 642. The student may supply 
any dimension that is omitted, taking such dimensions as he 
deems be^st. 

(977) Applying formula 233, 

^ = 4x4 = 1". Ans. 
4 

/ = g X 4 = -% say - ". Ans. 

(978) Applying formula 233, 

^ = 1 X 12 = 3". Ans. 
4 

^ = ^ X 12 =: 2". Ans, 
o 



MACHINE DESIGN. 323 

(979) Applying formula 235, 

^ = .2 X 2-^ -f .16 = .65", say — ". Ans. 
16 "^16 

/ = .1 X 2^ + .16 = .4% say ^". Ans. 

(980) Applying formula 237, 

^~ 8 +16-4- ^'''• 

76 +16 ~ 16 '^^y 8 • ^''^• 

(981) Applying formula 238, 



;f1-85 



</=5|/i:^ = 1.26'. 



Using formulas 236, 

^ = :^.^ = Ar, say -^^ Ans. 
6 . lb 

1.26„ 1,, . 
/ = — ^" = - '. Ans. 
5 4 

(982) Area of rod corresponding to diameter d^ = 
6^0 =^-^^^^^-^"- 



Hence, d = f^^^ = 1.22^ say 1 ]■". Ans. 
' .7854 "'4 

From formula 239, 
^. = ,816^,or^=-4 = M£ = i.53' = lg". Ans. 

The remaining dimensions are readily calculated by using 
formula 239. The dimension I? in Figs. 656 and 658 
should extend to the top of the cotter instead of to the top 
of the slot, as shown. The student may take d as the depth 
of the cotter. 

(983) Applying formula 241, 

./ 10,000 

'^^ ^-"^yu, 000X400 = ^' "^'^'-^y- ^°^- 

Hence, / = ^^ = 8. 33', say 8 |'. Ans. 



324 MACHINE DESIGN. 

(984) Applying formula 244, 

= 2% nearly. Ans. 



./ 15,000 

= i.06r- 



|/14,000X 1,200 

Using formula 245f 

15,000 _ 1, 
^-2 X 1,200- S- ^'''• 

(985) From formula 246, 

3/7 3/ 9 5 19 

^, = ^f^:zrl5|/ ^^ = 1.583% sayl3^". Ans. 

(986) Applying formula 251, 

d = ^/gi I l^^QQQ = 10.9% say 10 1% Ans. 
'^ ^ 15 X ^ X 6 8 

Since / = . 8 ^, and ^ = ^^^, ^ = :M_1^::. .4 (10 J _ 8) 

= 1.15", say 1-|". Ans. 

(987) Applying formula 249, second column of 

Table 47, 

^=.07 1/1,200 = 2.425", say 2—". Ans. 

4 

(988) {a) Apply formula 256. Here, m = -', hence, 



4 
(/^) Diameter of hole = 10.38* x x = 6", nearly. Ans. 

(c) Weight of solid shaft per foot of length = 

•^^^^^ ^^' X 490 = 267 lb., nearly. 
144 

Weight of hollow shaft per foot of length = 

:^«5£I(M)l^^X 490 = 191 lb., nearly. 
144 

Difference - 267 - 191 - 70 lb. per foot. Ans. 



MACHINE DESIGN. 325 

(989) {a) Using formula 208, 



/ = .24 4/3.254- .025 - .175 = .3", nearly. 
By formula 210, 

n = ^ = 3.333, say 3-^. 

Then, p z= — •= —- = — . Ans. 
{b) Using formula 212, 

p = -^ = lA\ say l|". Ans. 
(c) Using formula 215, 

/ = ?-^ = 1.6'^sayl|''. Ans. 

(990) {a) Using formula 21 for each case, 

?2 = 1 -^ y = 3-; ;2 = 1 --- 1- = — , and ?2 = 1 -J- 1- = — . Ans. 

{b) Using formula 211 for the first case, 

d, = d-^" = ?,. 25" - i^" = 2. 879". Ans. 

7t 3.0 

For the second case, it will be remembered that t = — /; 

lit 

hence, 

«r.r=^-2g/) = ^-/ = r-l|" = 5|'. Ans. 

3 3 

For the third case, t ^=z — t =^ — p -. hence, 

d =d- ^i^p) = ^ - 1.5/ = 12 - 1.5 X 1. 625 = 9^". Ans. 

(991) See Art. 1930. 

(992) See Art. 1935. Using formula 217, 

Referring to Table 43, this area lies between the areas 
corresponding to diameters of 2" and 2|"; choosing the 
larger, the outside diameter should be 2^". Ans. 



326 MACHINE DESIGN. 

(993) From Table 43, the effective area of a 4-inch bolt 
is 9.993 sq. in. Referring to Art. 1935, and using 
formula 217, 

W=aSt = 2X 9.993 X 8,000 = 159,888 lb., say 80 tons. 

Ans. 

(994) (^) Using formula 218, 

d= .02284/3,600= 1.368", say If". Ans. 
(<^) Using formula 219, 

n^ = .0052 X ' , = 9.89, say 10 threads. Ans. 
J., o i o 

(995) Applying formula 217, and taking 5^ as 6,000 
lb. per sq. in., 

3,200 __ 

^^Moo^-^'^^^^-^^- 

From Table 43, the nearest diameter is 1", 
Applying formulas 220, 221, 224, and 226, 

^ = 4xi"+f6'-ir6"- ^- 

^ = /.' = l"-^^"=J|". Ans. 

D^ = 2.12 X l"+.09" = 2.21". Ans. 
The student can readily make the drawing by referring 
to Fig. 605. Omit the washer. 

(996) The dimensions are readily obtained by aid of 
formula 229. 

(997) (a) Using formula 233 for all three cases, 

1117 

r^) ^=1x5 = 1^^ Ans. 

' ' 4 4 

/ = - X 5 = -", say —\ Ans. 



MACHINE DESIGN. 327 

t = ~ X 1^ = -:". Ans. 

0/5 4 

(998) Each key takes up 1,000 -^ 2 = 500 H. P. Hence, 
applying formula 232, 

, 18/^ 18 X 500 ^ ^. «;r A 

^=7Z77- ioxiox50 -^'^>^"y^- ^^^• 

/ = |^^|x2=l. 333", say l|^ Ans. 

(999) For the driven pulley, apply formula 234. 
-5 = ^' + ^' = .64', say I'. Ans. 

For the driving pulley, apply formula 235. 

^ = .2 x 4.5' + . 16" = 1.06", say 1^". Ans. 

16 

^ = .1 X 4.5'' + .16'' = .6r, say I". Ans. 

(1000) Using formulas 237, 

^ = '-^ + Te = -''''^^yi'- ^-- 

3 

(1001) It is evident that a 3-" shaft will transmit a 
^ ' 4 

great deal more than 6 H. P. at 135 R. P. M. Hence, using 

formulas 238 and 236, 

d = 5V^ = 1. 77". Therefore, 
loo 

^ = ^=.59^say^^ Ans. 

o o 

1 77 ^ 

t = ^ = , 354^ say ^". Ans. 
5 o 



328 MACHINE DESIGN. 

(1002) The calculation of d^ d^^ and the other dimen- 
sions is in all respects similar to that required in example 982. 
The various dimensions in the figure may be obtained by 
using formula 239. 

5 

(1003) According to Art. 1969, bt should equal - 

times the sectional area of the strap, and t =. — b. Hence, 

1 5 3 

bx jb^jy-Ty^ 3.5; or, b^ = 13.125 sq. in., and b = 

>|/13.125 = 3.623% say 3^". Ans. 

o 

, 1,1 ^5 39, 15, . 

The other dimensions are readily obtained from Fig. 661, 
and the drawing can be made by referring to it. 

(1004) Art. 1969 states that b may be made equal to 
^when a steel cotter is used in a wrought-iron rod. Hence, 

l, = d=2l". Ans. 

Also, • t = ]-d = l-x2l = ^'\ Ans. 
4 4 2 8 

(1005) From Table 43, the area at the bottom of the 
thread for a 3^" standard screw is 7.55 sq. in. Referring to 
Art. 1935, it would not be advisable to take 5^ greater than 
4,000 lb. per sq. in. in this case. Hence, W= aSt = 7.65 X 
4,000 = 30,200 lb. = load which rod may safely carry. 

When held by a cotter, the rod is weakened owing to the 
cutting out of the slot, the equivalent diameter being (see 
formula 239)<= .816^= .816 X 3.5 = 2.856"; thatis,arod 
having a diameter of 2.856" would be, theoretically, of the 
same strength as a rod having a slot in it for a cotter and 3^" 
in diameter. Hence, the equivalent area is 2.856^ X .7854 = 
6.41 sq. in. The area is less than in the first case, but the 

value of St may now be taken as — ^-r — = 5,500 lb. per sq. 



MACHINE DESIGN. 329 

in., and the safe load is 6.41 X 5,500 = 35,255 lb. Hence, the 
rod and cotter are, apparently, slightly stronger than the rod 
and nut. 

(1006) (^) Applying formula 240, 

d= 2.26 1/?4?^ X 1.5 = V. Ans. 

4,250 

/=ll^=ll-X3 = 4|". Ans. 

(^) Pressure per square inch of projected area = 

5,000 ^^^ .. . 
/ = ^ , , = 370 lb. Ans. 
^ 3 X 4.5 

(1007) Applying formula 240, 

^ = / = 2. 26 /^^ X 1 = 2. 45^ say %\\ Ans. 

(1008) See example in Art. 1975, and Table 46, Art. 
1978. 



^=1.5 ^ ^^-^QQ 1 = 3. 65^ say Z-'' . Ans. 
/I2,000x750 8 

Whence, /=^|^^= 6. 55\ say 6|". Ans. 
(1009) As in example 1008, 

^=1.5 \fZ^^^^ = 1. e'^, say ll''. Ans. 

4/8,500 X 275 8 

Whence, / = .3''^';,^^^ = 3. 92", say 4". Ans. 

P 

(lOlO) Here, /= 2^. From formula 242, /^ = -j. 
Hence, 2^X^=2 ^'r= ^^ ; or, ^= 1.846", say 1> 

7 3 ^"' 

Also, /= 3^=2 X 15- = 3-r''- Ans. 
(101 1) Circumference of wheel = 3.75)1; 
R p M - 7V^- ^-'^^Q >^ ^° - 407 



330 MACHINE DESIGN. 

^, ^ 220,000 ^,^., 

Then, / = ' = 540 lb. per sq. in. 

Applying formula 241 , and getting S^ from Table 46, 

^= 1.54/1^^^^ = 2.97^ say 3". Ans. 
i/l2,000Xo40 

Hence, / = i^i^ = 6. 1 7\ say 6^ ". Ans. 
3X540 ' ^ 4 

The other dimensions may be obtained from Fig. 665 and 
formula 247. 

(101 2) (a) Using formula 243 and Table 46, 



d=l. 13/1^ X2^= 2.09", say 2^\ Ans. 

/ = 2. 5 ^ = 2. 5 X 2^ = 5,4^ say 5^". Ans. 
o lo 4 

{3) From formula 242, 

P 9,600 ^^^ ,, 
^^Td"^ 21 X 5^ ^ P^^ ^^- ^^' • 

(1013) Using formula 250, 

^= .004|/7W3= .0044/25 X 3,000 = l.r, say 1-^ Ans. 



(1014) Using formula 249, 

^ = . 05 /P = . 05 /9^000 " 4. 74% say 4-?^ Ans. 

(1015) Applying formula 251, 



d = lA^^flliMr = 8.8% say S^\ Ans. 
' 15 7r X 4 ' -^ 4 

^ = i-K-^) = .^(8f-6) = l|". Ans. 
/ = .8 ^ = .8 X l|" = 1. 1\ say 1-^". Ans. 

o o 

Taking .y as 2 /, .y = 2 X 1-^' = 2j\ Ans. 

(See Fig. 668 for manner of representing the journal.) 



MACHINE DESIGN. 331 

(1016) This shaft should be calculated with the help of 
Table 49. To determine Z, we have, from the table of 
Bending Moments, 

B = W^ = 10 X ^4^ = isl tons. 
/ 8 4 

The twisting moment T = PR = 63,025 X -tt in. -lb. (see 

formula 231) = 

03,025 H, 63,025 300 .^.or 

2,000 X 12 ^ 77 ^^'-^"^^ = 2,000 X u ''-W = ^^-'' ^^-^""^- 

Then, Z= ^ = ^yj^^ 1.4, nearly. 

From Table 49, f^ = 1.461 for Z= 1.4. Hence (see 
Table 32, Art. 1416), 

< ^dsfk ^ 1.461 X 3.3 y ^=1.461 X 3.3^/^ = 8.25". 

Ans. 
(lOl 7) Applying formula 253, 

^=^^-^^rio^rW = '-'^ 

Z=:^=.75. 

From Table 49, y^ = 1.209 for Z= .6, and 1.277 for Z = 

.8. Hence, 1.277-1.209 : .8-.6 = ;ir : .75 -.6; or, ;f = .051. 

Therefore, forZ= .75, \/k = 1.209 + .051 = 1.26, and^^^ 

1.26 X 3.35 = 4.22', say 4-^^ Ans. 

4 

(1018) Use formula 256. In this case ;/2 = —. 

Hence, d=d^ s/\ - m' = 13y 1 - (^) = 12.8', say 12 ^^ 

Ans. 

(1019) Use formula 256. Here 7;^ = - =.5. Hence, 

3 V — 1 1 

^ = 167y -— ^, = 17.ir, say 17-x". Ans. 
* ^ 1 — .0 o 

rf, = 17^ -^ 2 = 8^". Ans. 



332 MACHINE DESIGN. 

(1020) The student should be able to draw the joint by 
referring to Art. 2000 and Fig. G76. The keys may be 
proportioned by formula 233. The student should use his 
own judgment regarding any dimensions not given. 

(1021) Fig. 673 and formula 258 are a sufficient guide 
for the drawing of this coupling. 

(1022) (^) From formula 260, 

n = ln-\-2 = lxll-\-2=^5l,S2iy 6. Ans. 

DO 6 

{b) From formula 259 and Table 50, 

^ = >^ Z> = . 239 X 11 = 2. 629", say 2|". Ans. 

o 

{c) From formula 260, 

^= Z> + 2 ^=11 + 2 X 2.625 = 16^". Ans. 

4 

{d) C = D -\- 4.25 ^ = 11 + 4.25 X 2.625 = 22.16, say 
22^". Ans. 

(1 023) Using formula 250, 

d= .0044/^77 = .004|/100X 600= .98", say 1". Ans. 



MACHINE DESIGN 

(QUESTIONS 1024-1063.) 



(1024) See Art. 2004. 

(1025) See Art. 2016. 

(1026) Radius of gear = 15" = 1^ ft. = R. From for^ 
mula 231, 

^ 63,025// 63,025 X 20 ^^^ ,^ 

^ = -1^- = 120 X 15 = ^^^ ^^- ^"^'^y* 

From formula 267, 

^ 9,600,000 9,600,000 o ^n/iK 

o = — , ^ ^^.^ = — — .^^ , c. -.r.^ = 3,094 lb. per sq. m., 

2^ + 2,160 2.5x^^x120 4-2,160 ' f ^ ^ 

nearly. 

Hence, from formula 266, 

A^ ,aoP 16.8X700 __ 
i.C=U.8^= 3,094 -3.8 sq.m. 

Assuming that d = 2-C, d C = %5 C = 3.S sq. in., or C = 

TT X 30 



y~ = 1.2329". Hence, number of teeth = 
2.5 



1.2329 



= 76 4-, say 76. Ans. 



Circular pitch = "" ^^^ = 1.2401". Ans. 

(1027) The velocity of a point on the pitch circle is 

2 
TT X 4- X 60 = 880 ft. per min., nearly. Hence, from for- 
o 

mula 267, 

For notice of copyright, see page immediately following the title page. 



334 MACHINE DESIGN. 



c, 9,600,000 o-i^oiu • 1 

~ 880 + 2 160 ^ ^^^ ^^' ^^'^ ^^^^^y* 



From formula 231, 



^ 63,025 X 300 ,, _.. ^ ,, 
^= 60X28 =l^>^54.5lb. ^ 

From formula 266, 

,^ 16.8X11,254.5 _^^ 
^^= 305^ =59.87 sq.m. 

Taking ^ equal to 3 C^ d C^ •=■ 59.87 sq. in. ; whence, C = 
4.4673". 

r X 56 

Number of teeth = f-j^ = 39.4, say 39 teeth. 

4.4673 '' 

rryu r 3.1416 X 56 , ^^^„ . 
Then, C— — = 4.511^ Ans. 

Breadth of face = 

<^ = 3 (7= 3 X 4.511 = 13.533, say V^\ Ans. 

Thickness of tooth = .475 (f= .475 X 4.511 = 2.143". Ans. 

Height above pitch line = .3 6'=. 3 X 4.511 = 1.353". Ans. 

Depth below pitch line =.4(7 = .4 X 4.511 = 1.804". Ans. 
Using formula 123, Art. 1415,- 

diameter of shaft = ^ = 4.92 y ^ = 7.36", say 7|". Ans. 

60 o 

The enlargement for the wheel seat will have a diameter 

of 1,2^= 1.2 X 7q = 8.85", say ^-"' Ans. 
o o 

By formula 273, thickness of hub = ■ 



w-\ VbCR = J ^13.5 X 4.511 X 28 = 3. 983", say 4". Ans. 
The length of the hub may be 

1. 4 ^ = 1, 4 X 13^ = 18. 9", say 19". Ans. 

Formula 272 gives for the number of arms, 

-8r = . 55^97^ = .55^/39' X 4.5U = 5 arms. Ans, 



MACHINE DESIGN. 335 

By formula 270, width of arm at center of wheel = 

^ o 15 

Assuming the taper to be — on each side, the width of 
arm at pitch line is 

B^ - (2 X ^ X 28) = l-^\ nearly. Ans. 

Thickness of arm = ]- C =\ X 4.511 = 2.255" = 2^. 

Ans. 
Thickness of stiffening rib = .4 (7= .4 x 4.511 — 1,804". 

Ans. 

(1028) By formula 275, 

• p= 2/= 280 X i = ^^. 

Radius of wheel = R= ^ = 36". 

2 

Then, by formula 231, 

rr PRN 280X36X110 ^ o^. TJ r» a 
^^-63:025^ 3X63,025 = ^-^^^ H. P. Ans. 

(1029) See Art 2049. 

(1030) {a) By formula 282, 

{b) By formula 283, w = 1.43 D' = 1.43 X 1* = 1.43 lb. 
Ans. 

(1031) By formula 291, 

P= 6,600^' = 6,600 X 1.25' = 10,312.5 lb. Ans. 

(1032) See Art. 2005. 

(1033) The values of the different dimensions, as 
determined from the table of proportions in Art. 2014, 
are as follows; 



336 MACHINE DESIGN. 

a =14 + 1 = 15^ p =.25 X 14 + .625 = 41". 

/^ = .5 X 14 + 1 = 8^ ^ = 1.75 X 14 = 24^". 

c = .66 X 14 = 9.24", say q' = 1.5 X 14 = 21". 

9i". r = . 15 X 14 = 2. 1", say 2^'. 

e =.825x14-. 25 = 11. 3", r' = .1 x 14 = 1.4", say If". 

say \\^\ r, = 14". 

/ = .6X14=8.4", say SJI". s = .9 X 14= 12.6", say 12|". 

^ =.1 X 14 + . 5625= / = .15X 14 +.375 = 2.475", 

1.9625' say l|i". say 2^". 

h = .1 X 14 + .25 = 1.65^ /'=. 9 X 14=12.6", say 12|". 

say lfi"» 2^ = 1.5 X 14 = 21". 

h = ,08 X 14 = 1.12", say e' = .25 X 14 + .375 = 3F- 

li". 2£/ = 1.45 X 14 = 20.3", say 

i = .11 X 14 = 1.54", say 20i". 

1^". w'^lAl X 14 = 20.58", say 

j = V' ^of ". 

k = .5 X 14 + 1.25 = 8i^ w^ = 1.75 X 14 = 24V'. 

/ = I". jir = .1 X 14 = 1.4", say If", 

m = .175 X 14 + .3125 = _^ = .3 X 14 + .75 = 4.95", 

2.7625", say 2f". say 5". 

n = .25 X 14 + .25 = 3f". / = .2 X 14 + .5 = 3.3", say 

n' =.lx 14+.375 = 1.775", 3i". 

say If". z =.09X14=1.26", say li". 

o = 1". ^' = 2i". 

The student may use his own judgment to supply any 
dimensions not given. 

(1034) See Art. 2023. 

(1035) Number of teeth = A^= 56 X Ij = "^0. 

Circular pitch = C = ^illii = 2. 5133". 

I4 
Applying formula 272, 



^=.55 4/iV^ = .55|/70' X 2.5133 = 5.79, or 6 arms. An& 

(1036) See Art, 2037. 

(1037) See Art. 2057. 



u.^ 



MACHINE DESIGN. 337 

(1038) See Art. 2066. 

(1039) SeeArt. 2074. 

(1040) See Arts. 2005 and 2006. 

(1041) SeeArt. 2022. 

(1042) The values of the different dimensions, as de- 
termined from the table of proportions in Art. 2011, are: 

d=.^". 
a = 3.25 X 3.5 = llf". n = 1.25 X 3.5 = 4|". 

d =1.75X 3.5 = 6i^ . o = Y, 

^ = .5X3.5 = 11". ^=.625 X 3.5 = 2^^ 

/ = . 4375 X 3. 5 = 1. 53125^ r = .25 X 3. 5 = f'. 

say lif. s = .1875 X 3.5 = fV, say 



^ = .09 X 3. 5 = .315% say 




1 \'l 

1 6 • 






5 f 
1 6 • 


/ = 


.65X 3.5 = 2.275", 


say 




/z =.3125 X 3.5 = 1-//. 




9 5^ 






t = .25 X 3.5 = F. 


u = 


.75X 3.5 = 2f". 






J =.375 X 3.5 = lyV- 


V = 


1.375 X 3.5 = 4f|", 


say 




k = 1.0625 X 3.5 = 3.72", 




4r. 






say 3|". 


X = 


.25 X 3.5 = i". 






/ =.875 X 3.5 = 3iV". 


y = 


.5 X 3.5= If". 






m=: 1.75 X 3.5 = 6^". 


z = 


.0625X3.5 = -^", sayi^. 





(1043) See Arts. 2020 and 2018. 

(1044) Applying formula 268, 

^ = ^i = ^"^^ = ^-148", say 2^". Ans. 

(1045) Applying formula 274, 

,- ^A _ 4 X. 5236 _.g.o^ . 5 
^ - TTaT^ - ^^^5236+1? " ^* 

(1046) See Art. 2038. 

(1047) Applying formula 276, 

,_ ^ + ^ , l._ lQ + 52 1 _ 3, 

^-"^00~ + 16 -^0^+16--^^^^''^^8 • ^'''• 
7). 0. III.— 35 



338 MACHINE DESIGN. 

(1048) Applying formula 283, 

^ = 420 -^ 2 = 210. 
Applying formula 285, 



1.0941b. 



k 



2w ^ 4w' 2 



2,023 



-/ 



2,023^ 



210' 



2X1.094 ' 4X1.094 

(1049) See Art. 2047. 

(1050) See Art. 2050. 

(1051) Velocity = 

Using formula 281, 



12 ft. Ans. 



18 X TT X 72 



H = 



vD' 



{^'' - ife ) 



60 
67.86 X (li) 



67.86 feet per second. 



Y2OO 



825 V 107.2/ 825 

20. 18 H. P. 20. 18 X 24 = 484. 32 H. P. Ans. 



67.86' 
107.2 



) 



(1052) By formula 283, 

ze; = 1.43 /?' = 1.43 X (l^) = 2.23 lb. per foot. 

^ = 386 -^ 2 = 193 ft. 
Applying formula 284, 

r- !^ 4. ^^ - A?iAi95!.+ 2.23 X 9.25 = 4,511 lb. 
~ 2^ -t-^/^- 2X9.25 ^ ^^g 

(1053) See Art. 2070, 



n = .1S^P= . 134/5,600 = 2.31, say 2. 



d= .0115l/^ = .0115|/-54^ = .7237", say l". 
yn 4/2 4 

^' = 1.2^=L2x^ = .9^say||^ 

4 ^32 

Remaining dimensions are the same as those calculated 
in Art. 2070. 



MACHINE DESIGN. 339 

(1054) The values of the dimensions given in the 
table of proportions in Art. 2019 are; 

^ = 3 X 2.5 = 7^". m = .Vlh%%.h = ^\ 

b =^l.bx 2.5 = 18|^ n = .25 X 2.5 = f\ 

c = 5.375 X 2.5 = 13i^". ^ = .1875 X 2.5 = i|",say i". 

^ = 3 X 2.5 = 7Y. P = .375 X 2.5 = H'/say 1^ 

/= 1.75 X 2.5 = 4f^ ^ = .625 X 2.5 ^lyV- 

^ = 1.5 X 2.5 = 3|". r = 1.25 X 2.5 = 3f'. 

A =2.125 X 2.5=5^''. .y = .25 X 2.5 + .5 = 1^". 
i =.16x2.5=.4",say^". 2; = 1.125 X 2.5 + .1875 = 3*. 

J = .25 X 2.5 = 1". w = 1.5 X 2.5 = 31". 

^ = 1.875 X 2.5 = 4H^ 7 = .0625 X 2.5 = j%". 

/= 1.125 X 2.5 = 2i|". 2 =2.5,- .0625 = 2^\ 
For the other dimensions necessary to draw the bracket, 
see Fig. 697. 

(1055) Using formula 271, 

^?=1.75j/f^^==4.15^say4'. Ans. 
1^ X 6 ' -^ 8 

-.= -X4-=2-^ Ans. 

(1056) (a) The diameter of the wire composing the, 
rope is, by formula 282, 

^=^ = 1=1^ 
9 9 12 

The weight of the rope per foot is, by formula 283, 
w = 1.43 D' = 1.43 X (I) = .8044 lb. 

Stress due to bending is, by formula 287, 

^ £^d 25,000,000 X t'i -.oo^^iu 

-^^ = 2^ = 13X12 = '''^'' ^^' P^^ ^^- ^^- 

Stress due to centrifugal force is, by formula 238, 
^ , 4x.8044x(^°y 



340 MACHINE DESIGN. 

Hence, S, = S - {S, ~j- 5,) = 25,000 - (13.355 + 758) = 
10,887 lb. per sq. in. 

T^ = maximum tension on driving side = 

^d'nS, = ^X Q-S X 42 X 10,887 = 2,494 lb. 

T; = i r, = 2,494 X .5 = 1,247 lb. 

P= driving force = T^- T.^ = 2,494 - 1,247 = 1,247 lb. 

„ rr PV 1,24 7 X 5,000 ^__^ ^ 

Horsepower = H= -33^ = 33,000 = ^^' ^' ^' 

(d) The greatest deflection will, of course, be on the 
driven, or slack, side. Hence, applying formula 285, 

, _ T^ J~T^ a' _^ 1,247 . /" 1,247^ "21275 _ 

'~ ^w ^ 4:W^ 2~2X.8044 '^ 4 X •8044'* 2 ~ 

14.7 ft. Ans. 

(1057) Circular pitch = 0="^= ^^4r^ = •'^854^ 
^ ^ ^ 84 84 • 

Applying formula 272, 

^ = .55 ^84' X .7854 = 4.74 +, say 5 arms. Ans. 

(1058) Using formula 277, 

//'BR //8 X 11 , o// « "^/z A 

a = y ■ = y — = 2.8'', say 2-". Ans. 

n 4: 8 

Assuming the taper to be — on each side, the taper 

for both sides will be -— X 2 = r-r. The width of the arm at 

48 24 

rim will be 

2^ - (^ X 11) = 2.417", say 2|". Ans. 

7 1 7 
Thickness at center =z 2- X ^ = ^TF-"- Ans. 

o /I Id 

3 1 3 

Thickness at rim = 2- X ^ = 1^7/'- Ans. 

8 2 lo 

(1059) Using formula 279, 

S' = A1 = L>^ = 2.406 sq. in. 
4 4 



MACHINE DESIGN. 341 

Net section of one bolt = 2.40G ^ 6 = .401 sq. in. 

7 
Hence, according to Table 43, —" in. bolts will be used. 

Ans. 

(1060) (1) The dimensions as taken from Table 52, 
are: 

G = |';i7=l^',and/=|'. 

(2) From the table of proportions in Art. 202 1 f 

^? = 2 X 3 = 6'. y = .4 X 3 = 1.2^ say \\\ 

^=1.5 X 3 = 4i^ k = .3x3 = .9^ say if or l''. 

^ = .25 X 3 + .375 = li''. / =.15x3 = .45",say-i|"orV'. 

e = 1.25 X 3 = 3f". m = 1.5 X 3 = U". 

/=. 2x3+. 125 = . 725, say f^ 7i == .2x 3+.25 = .85", say }". 
g= 1.75 X 3 = 5i". ^ rr .15 X 3 = .45", say \" . 

h= 1.4x3 = 4.2", say 4i'. 

(3) The various dimensions are readily obtained from 
Fig. 735. 

(4) The dimensions, as obtained from Table 54, are: 

^ = 12"; ^ = .79"; ^ = 1" \ ^= 3|"; ;^ = 12"; ^ = 1^';/ = 
loi", and^=19^ 

( 1 061 ) Applying formula 280, 
w=.3Z>' = .3 X (l|) =.91875 lb. per ft. 

Hence, total weight = 875 X .91875 = 803.9 lb. Ans. 

(1062) {a) Using formula 281, 



»-"xM, 



(■I) 



^-825^^^ 107:2 ;-■ 825 V 60^X107.2;- 

38.223 H. P. Ans. 

{b) Referring to the diagram, Fig. 723, the horsepower 
appears to be about 38 H. P. Ans. 



342 MACHINE DESIGN. 

(1063) Applying the method described in Art. 2063, 
and assuming that -7 = 850, we proceed as follows: 

The driving force = P= 2M22_^ ^ ^^^^^ = 842 
lb«, nearly. 

Then, T^ = 842 X 2 = 1,684 lb. 
Stress due to bending is 

Hence, 5, = 25,000 - 14,706 = 10,294. 

r 1 

Cross-section of wires z=z -^ = — t: d^ n, or 

^ Tt nSt '^ TT X 42 X 10,294 
The diameter of the rope is, by formula 282, 

D=^^d=^X .0704 = .6336, say |". 

o 

A recalculation taking into account the stress due to cen- 
trifugal force will not appreciably affect the result; hence, 

5 
the diameter of the rope is — ". Ans. {c) 

o 

(a) Diameter of wires is, then, 

^-^9 = .625-^9 = .0694^ Ans. 
o 

{b) Radius of pulley = .0694 X 850 = 58.99", say 59^ 

Diameter of pulley = 59x2= 118" = 9 ft. 10", say 10 ft 

Ans. 



INDEX 



A 






PAGE 


Acceleration 321 


Accelerating force . 






321 


Action, Arc of . 






908 


" Angle of 






908 


Addendum line , 






902 


" circle 






902 


of tooth . 






902 


Aeriform bodies 






298 


American thread 






1224 


Angle of action . 






908 


" of approach . 






908 


" of obliquity . 






914 


" of recess . 






908 


Annular gears . 






911 


Approach, Arc of 






908 


" Angle of . 






908 


Approximation, Tredgold 


's 




918 


Arc of action 






908 


" of approach. 






908 


" of recess 






908 


Arms of pulleys 






1334 


" of wheels . 






I319 


Atom .... 






297 


Axis of beam, Neutral 






787 


B PAGE 


Back gears 875 


" gears, Differential 




• 895 


" lash . 




• 903 


Balancing Pulleys . 






1342 


Ball and socket bearings 






• 1307 


Base circles 






914 


Beams .... 




77 


1,813 


" bending moment 






775 


" Cantilever 






771 


" Comparison of strength anc 


1 


stiffness 


• 797 


" Conditions of equilibrium 


• 772 


" Continuous . 






. 77t 



Beams, Deflection of. 

" Experimental law of 
" Neutral axis of 
" Restrained . 
" Simple . 
" Vertical shear in . 
Bearing surface of journals 
Bearings .... 

" and slides. Pressure on 
" Ball and socket . 
'' Distance between 
*' Divided 
" Footstep or pivot 
"■ Solid journal 
Bell crank lever 
Belt fastenings . 
" Power transmission by 
" pulleys 

" pulleys, Counterbalance f 
" pulleys, Examples of 
" Quarter turn 
" shifting mechanism . 
" to connect non-parallel shafts 
" To determine horsepower of 
" To determine width of 
" transmission, Examples of 
Belting, General rule for 
" Rope 

" Use and care of 
Belts .... 
" Climbing of 
" Double 
" Guiding , 
" Open and crossed 
Bevel gears 

" gears, Proportions of 
Block, Pulley . 
Blow, p-orce of a 
Bodies, Composition of 



PAGE 

794 
789 
787 
771 
771 

773 
1263 
1293 
1265 
1307 
1275 
1294 

1313 
1293 

817 

865 

852 

1332 

1342 

1342 

867 

885 

866 

855 

853 

870 

8S5 

1343 

863 

1332 
864 
856 
864 

851 
917 
1325 
341 
357 
297 



VI 



INDEX 









PAGE 


Body, Gaseous 2q8 


"■ Liquid 






. 298 


" Solid 






. 298 


Bolt for stone work . 






. 1238 


" heads . 






• 1236 


" Patch . 






. 1243 


" Stud . 






. 1242 


" Tap 






. 1242 


Bolts .... 






1224, 1242 


" Foundation 






• 1238 


" in shear 






. 1244 


" Proportions of . 






• 1234 


" Strength of 






• 1233 


Box coupling 






. 1281 


Brackets, Wall . 






• 1309 


Brake, Prony 






. 1651 


Brass .... 






. 1223 


Brittleness . 






• 300 


Bronze 






. 1222 


" Manganese 






. 1223 


" Phosphor 






. 1223 


Built-up wheels . 






. 1323 


Button-head screws 






• 1243 


C 'page 


Cam, Harmonic motion . . . 844 


" Plate . 






839 


" Positive 








845 


Cams . . , 








839 


Cantilever . 








771 


Castings, Chilled 








1222 


Steel . 








1222 


Cast iron 






741 


, 1219 


" iron, Malleable 








743 


" iron, Shrinkage 


of 






1220 


Center, Dead 








821 


" line of motio 


n 






813 


" of gravity 








330 


" of moments 








329 


Centiifugal force 








327 


Centripetal force 








328 


Chain drums 








1360 


Chains 






. Bog 


. 1359 


Flat link 








1361 


*' Strength of 








1359 


Chilled castings 








1222 


Clamp coupling . 








1282 


Claw coupling . 








1288 


Clearance, Tooth 








903 


Click . 








923 


Climbing of belts 








864 


Clutches, Friction 








1289 


"■ Shifting gear f< 


)V 




1290 


Clutch gearing . 






883 


Coefficient of elasticity 






751 


Collar journals , 


. 






. 1270 



PAGE 

Columns 798 

Combination of pulleys , . . 342 
Composition of forces . . , 307 
" of moments . . 766 

Compound lever .... 337 
Compressibility .... 300 

Compressive strength . . . 753 

Concave key 1246 

Condenser, Cooling surface re- 
quired by 734 

Conditions of equilibrium . . 334 

Cones for crossed belts . . . 859 

" for open belts . . .861 

Connecting rod 813 

" rod, Crank and . . 820 

Conservation of energy . . . 357 

Contact, Law of tooth . . . 907 

" Path of .... 908 

" Point of . . . .908 

Continuous beams .... 771 

Copper 1222 

Cotters 1253 

" Locking arrangements for . 1260 
" Strength of . . . . 1256 
" Taper of .... 1259 

Cotton belts 1332 

Countersink head bolts . . . 1237 
" head screws . . . 1243 

Couple 329 

Coupling, Box or muff . . . 1281 
" Clamp . . . . 1282 

" Flange .... 1283 

" Flexible . . . . 1287 

" Friction .... 1289 

" Loose .... 1287 

" Seller's cone , . . 1286 

Couplings 1281 

Crank 813 

" and connecting-rod . . 820 

" and slotted crosshead . . 822 

Crank-shaft pedestals . . . 1301 

Crossed belts 851 

" belts, Continuous cones for 859 
Crucible steel . . ... . 746 

Cycle of motions .... 812 

Cylinders and pipes, Strength of . 759 

D PAGE 

Dead centers 821 

'' points ..... 821 

Deflection of beams .... 794 

Density 358 

Diagram, Skeleton .... 827 

Differential back gears . . . 895 

" bevel train . . . 893 

" gearing .... 887 



INDEX 



Vll 



Differential motion 
" pullev 

Direction, Line of 

" of rotation 

Divided pitch 
Divisibility 
Double belts 
Double-threaded screws 
Driver and follower 
DuctiiitN' 
Dynamics . 

" of machinery 

E 

Eccentric rod 

Effective pull of belt. 

Efficiency . 

Elasticity 

" Coefficient of 

Elastic limit 

Elongation, Ultimate 

End journals 

Energ}', Conservation of 
" Kinetic 
Potential 

Engine, lathe train . 

Epicyclic trains . 

Epicycloidal tooth sj'-stem 

Equilibrium 

*' Conditions o 

" of beams 

" polygon 

Static . 

Expansibility 

Extension . 



Factor of safety 

Falling bodies . 

" bodies, Formulas for 

Fastenings, Belt 

Fillister head screws 

Fixed pulley 

Flanges, Pipe 
" pulleys 
" coupling 

Flanged nut 

Flat key 

Flexible couplings 

Follower, Driver and 

Foot-pound 

Foot-pounds, Bending moment ex 
pressed in 776 

Footstep bearings .... 1313 

Force ....... 303 

" Accelerating . . . 303, 321 
" Centrifugal . . . .327 



:i34' 



PAGE 

896 
343 
334 
851 

1231 
299 
856 

1230 
812 
301 

3"4. 317 
811 

PAGE 

813 
852 

351 
300 

751 

750 

752 

1262 

357 

354 
356 
875 



334 
767, 772 
772 
763 
329 
300 

299 



PAGE 

756 
320 

323 
865 

1243 

341 

1365 

1334 

1283 

1239 

1246 

1287 

812 

352 









PAGE 


Force, 


closed mechanisms 


• 845 


" 


diagram . 




763 


(. 


of a blow 




357 


ti 


of gravity 




318 


Cl 


Retarding 


303, 321 


" 


Unit of . . . 




30A 


Forces 


, Components of . 




316 


" 


Composition of 




307 


" 


in link mechanisms 




825 


" 


Parallelogram of 




309 


" 


Polygon of . 




312 


" 


Representation of 




307 


" 


Resolution of 




315 


"■ 


Resultant of . 




308 


" 


Triangle of . 




311 


Foundation bolts 




1238 


Friction clutches or couplings 




1289 


'• 


Coefficient of 




348 


'• 


gearing 




• 1330 


" 


Laws of 




• 34Q 


" 


of journals . 




12^^.8 


"• 


of worm gearing 




1327 


Fulcrum .... 




. 335 



G 

Gaseous body 
Gaskets .... 
Gas, Permanent . . . ' 
Gas-pipe threads 
Gear blanks. Sizing . 
" Positive 

" teeth, Proportions for 
" teeth, Shrouded 
" teeth. Strength of . 
" teeth, Wear of . 
" wheels 

" wheels, Proportions of 
" wheels, Hubs and arms of 
Gearing, Belt 
*' Clutch 

" Differential 

" Friction 
" Mangle 

" Materials for 

Tooth . 
'» Wire rope . 

*' Worm . 

" Worm or .screw 

Gears, Annular or internal 
Back 
" Bevel 
*' Single-curve . 
" Standard 
General properties of matter 
(Governors, Hartford water wheel 
Gravitation .... 



PAGE 

298 

1367 

29S 

1231 

904 

845 

903 

1317 

1315 

1317 

898 

1318 

1322 

1332 

883 

887 

1330 

881 

1314 

1314 

1350 

1326 

920 

911 

875 
917 
914 
914 
298 
897 
317 



Vlll 



INDEX 



Gravitation, Law of . 
Gravit}', Center of . 

*' Force of 

" problems, Formulas for 

" Specific 
Guide pulleys 
Guiding- belts 
Gun metal .... 
Gyration, Radius of . 

H 

Hangers .... 

Post .... 

Hardness 

Harmonic motion 

" motion, cams . 

Hartford water-wheel governor 
Heads, Bolt .... 

Hob for worm gearing 



PAGE 

318 

33^- 
318 

319 

358 

863, 1349 



PAGE 

Journal-box for small shafts . . 1208 



" (worm gearing) 
Hollow shafts 
Hook bolt . 
Hooks . 
Horsepower 

" of belts 

Hubs of gear-wheels 

" of pulleys 
Hydrokinetics . 
Hydrostatics 



Ideal twisting moment 
Idlers .... 
Idle wheels 
Impenetrability 
Inclined plane 
Indestructibility , • 
Inertia 

" Moment of 
Initial velocity . 
Intercept 

Interchangeable wheels 
Interference 

Internal or annular gears 
Involute gears . 

" rack 
Iron, Cast . 

" Malleable cast . 

" Pig . . . 

" Wrought . 



Joint, Double universal 

" Knuckle . 

u Toggle . 

" Universal 
Journal-box 



1222 

790 

PAGE 

• 131 1 
. 1309 

. 300 
. 823 
■ 844 
. 897 
. 1236 
921 

• 1327 
. 1280 

• 1237 
. 1363 

• 354 
. 855 
. 1322 

. 1336 

• 304 
. 304 

PAGE 
. 1278 
. 1349 
. 873 
. 299 

• 344 
. 300 

299. 305 

• 790 
. 324 
. 768 

• 909 

• 9'5 
911, 916 

912 

. 916 

741, 1219 

• 743 
. 741 

743, 1222 

PAGE 
. 836 

. 1244 

. 824 

835, 1287 

. 1297 



J V»Lll lldlO ..... 

" Bearing surface of . 


. 1201 
• 1263 


Collar .... 


. 1269 


End .... 


. 1262 


" Friction of . 


. 1268 


Neck .... 


. 1266 


" Pivot .... 


. 1269 


K 


PAGE 


Keys 


• 1245 


" Proportions of . 


. 1249 


Key-way 


. 1246 


Kinematics 


. 811 


Kinetic energy .... 


• 354 


Kinetics 


• 304 


Knee-joint 


. 874 


Knuckle joint .... 


. 1244 


L 


PAGE 


Lathe train .... 


. 875 


Law of gravitation . 


. 318 


Laws of friction 


• 349 


" of motion, Newton's 


• 304 


" of weight .... 


. 318 


Leather belts . . . , 


• 1332 


Lever 3- 


>5, 813 


Levers 


. 813 


" Non-reversing 


. 816 


" Reversing 


. 814 


" Bent .... 


• 337 


" Compound 


• 337 


Line of direction . . , 


. 334 


" shafting .... 


. 1271 


Link mechanisms 


. 812 


Liquid body . ._ . . 


. 298 


Locknuts 


1240 


Locking arrangements for cotters 


. 1260 


Loose couplings 


. 1287 


" pulleys .... 


" 1336 


M 


PAGE 


Machine 


811 


" construction, Materials 


used in . 


1219 


Machinery, Dynamics of . 


811 


Machines, Simple 


• 335 


Malleability .... 


300 


Malleable cast iron . 


743 


Manganese bronze . 


1223 


Mangle gearing .... 


881 


Mass of a body .... 


318 


Materials of gearing . 


1314 


" used in machine construe 




tion 


1219 


Matter 


297 


Mechanics 


301 


Mechanism ..... 


S12 



INDEX 



IX 





PAGE 




PAGE 


Mechanism, F'orce closed . 


. 845 


Path of a body .... 


. 301 


" Positive motion . 


• 845 


" of contact .... 


. 908 


" Reversing 


. 880 


Pawl 


• 923 


Shifting: belt . 


. 885 


Pedestals 


• 1297 


" slow motion . 


. 823 


" Crank-shaft 


• 1 301 


Mobility . . . . . 


299 


Phosphor-bronze 


. 1223 


Molecule 


• 297 


Pillar 


• 798 


Moment of inertia 


. 790 


Pillow-block .... 


. T297 


" Resisting 


. 793 


Pin keys 


. 1248 


" Twisting 


. 826 


" Split 


. 1260 


Moments, Bending . 


• 775 


Pipe flanges .... 


. 1365 


" bending. Diagram of 


777 


threads .... 


I23I 


"• Center of . 


329 


Pipes and cylinders, Strength of 


• 759 


" Composition of 


766 


Pitch circle 


900 


'' Graphical expressions fo 


r 767 


" Circular .... 


. 901 


" of forces . 


825 


" cylinder .... 


. goo 


" of inertia, Polar 


1273 


" diameter .... 


900 


" of inertia, Rectangular 


1273 


" Diametral 


. 901 


" Principle of 


826 


" line 


900 


Motion 


301 


" Normal .... 


• 915 


" Center line of 


813 


" of a screw 


347 


" Harmonic 


823 


" of screw .... 


1224 


" Newton's laws of . 


304 


" point .... 


• 937 


Parallel .... 


837 


Pitman 


. 813 


" Straight line . 


837 


Pivot bearings .... 


1313 


" Vibrating link 


829 


" journals .... 


1269 


" Whitvvortli quick-return 


833 


Pivots with loose disks 


I3M 


Motions, Cycle of . . . 


812 


Plane, Inclined .... 


344 


" Quick-return 


829 


Plate cam 


839 


Movable pulley' .... 


341 


Point of contact .... 


goS 


Muff coupling .... 


1281 


" Pitch 


9^7 


Mule pulleys .... 


870 


Polar moment of inertia . 


"273 


Multiple-tUreaded screws 


1230 


Pole distance .... 


768 


oS" 


PAGE 


" of a force polygon . 


764 


Naves of gear-wheels 


1322 


Polygon of forces 


312 


Neck journals .... 


1266 


Porosity 


299 


Neutral axis of a beam . 


787 


Positive motion .... 


845 






" motion cams 


845 


" equilibrium 


334 






' 




Post hangers .... 


1309 


Newton's laws of motion . 


304 






Nicholson's hydrometer . 


380 


Potential energy 

Power 


356 
353 


Non-reversing levers 


816 










" arm of lever . 


336 


Normal pitch .... 


915 


" transmission bv belt . 


852 


Nuts 


1224 










" Unit of . 


353 


" Forms of . 


1239 


Pressure on bearings and slides 


1265 


" Locking .... 


1239 


Principle of moments . * . 


329 


" Proportions of . 


'234 


Projectiles 


325 


O 


PAGE 


Properties of matter . . . . 


298 


Obliquity, Angle of . 


914 


Proportions for gear-teeth 


903 


Open and crossed belts . 


851 


Pull of a belt, Effective . 


852 


" hearth steel . . . . 


746 


Pulley . 


341 


Outline, Tooth .... 


908 


" Differential . . . . 


343 


P 


PAGE 


Hub of 


1336 


Parallel motion .... 


837 


Pulley, Loose 


1336 


" rod . . . . . 


813 


" Rim of . . '. . 


1333 


Patch bolt 


1243 


" Split cast-iron 


1332 



INDEX 



Pulleys, Arms of 
Belt 



" Belt, Examples of 

" Balancing" . 

" Combination of . 

*' Flange . 

" for rope gearing 

" Guide . 

" Maximiim speed of 

" Mule . 

" Stepped 

" Tension 

" Tightening or guide 

" Wire rope . 

" Wrought-iron 

Q 

Quarter-turn belts . 
Quick-return motion, Whitworth 
" return motions 

R 

Rack and wheel 
Radius of gyration 
Range of a projectile 
Ratchet and screw 

" wheels . 
Ratio, Velocity . . . 348, 8 
Rays of a force polygon 
Reactions of supports 
Recess, Angle of 
" Arc of . 
Resisting moment 
Restrained beam 
Resultant force . 
Retarding force ... 30 

Reversible clicks 
Reversing levers 

" mechanisms 

Right-handed rotation 
Rims of pulleys . 

" of gear-wheels . 
Rocker .... 

Root circle .... 

" line .... 

" of a tooth . 
Rope beltiftg 

" gearing, Pulleys for 

" gearing. Wire . 

" Stresses in wire 

Ropes 

" Power transmitted by 
" Strength of . 
Rotating pieces . 
Rotation, Direction of 

" Right-handed . 

'' Left-handed 



PAGE 

1334 
1332 
1342 
1342 
342 
1334 
1347 
1349 
1338 
870 
862 

134Q 
863 

1357 
1337 

PAGE 

867 

833 
829 

PAGE 

QIO 
790 
325 

g26 
923 
2, 849 
764 
771 



807 



Round keys 
Rubber belts 



S 



7QO 

771 
308 

, 321 
924 
814 
880 
812 

1:333 
I3I8 

813 

902 

902 

go2 

1343 

13-17 

1350 

1352 

Ij45 

1345 

807 

1261 

, 872 

812 

812 



Safet}', Factor of 

Screw ..... 

" cutting 
gearing . 

'^ Pitch of . 

" Ratchet and 

" thread 

" threads 
Screws .... 

" Multiple-threaded 
Seats for bearings 

" for bearings, Lining for 
Seller's cone coupling 

" triangular screw threa 
Set screws . 
Shaft couplings . 
Shafting, Line . 

" Line, Speed of 

Shafts .... 

" at right angles 

" Hollow . 

" subjected to bending 

" subjected to twisting 
bending 
Shear . 

axis . 

" diagram 

" line . 

" Vertical 
Shifting-belt mechanism . 
" geai' for clutches 
" gear for couplings 
Shrink rule. 
Shrinkage of cast iron 
Shrouded gears . 
Simple beam 
Single belts . ... 

" -curve gears . 
Skeleton diagram 
Slides, Pressure on . 
Sliding keys 

Slotted crosshead, Crank and 
Slow-motion mechanism . 
Solid body .... 
Special properties of matter 
Specific gravity . 
Speed cones 

"■ of line shafting 
Split cast-iron pulley 

" pins .... 
Sprocket wheel . 
Spur gears, Proportions of 



PAGE 
. 1248 
• 1332 

PAGE 
756 
347 
877 
920 

347 
926 

347 
1224 
1224 
1230 

1295 
1297 
1286 
1224 
1244 
1281 
1271 
1274 
1271 
867 
1280 
1275 



and 



1278 

755 

775 

775 

775 

773 

885 

1290 

1290 

1220 

1220 

1317 
771 

1332 

914 

827 

1265 

1248 

822 

823 

298 

299 

358 

858 

•274 

1332 

1260 

1362 

1318 



INDEX 



XI 



PAGE 
. 1227 

• 334 

• 329 
304. 329 

744, 1222 

- 746 



Square threads . 
Stable equilibrium . 
Static equilibrium . 
Statics .... 

Steel . . 
" Bessemer . 

" castings 1222 

" Crucible 746 

" Open-hearth .... 746 
" Tempering and hardening . 745 

Stepped pulleys 862 

Steps for bearings .... 1295 
Stop jilate for locking nuts . . 1241 
Straight-line motion .... 837 

Strain 748 

Strength and stiffness of beam . 797 
*' of chains . . . 809, 1359 

" of cotters .... 1256 

" of gear-teeth . . . 1315 

" of keys .... 1249 

" of screw bolts . . . 1233 

" Ultimxate .... 752 

Stress ....... 748 

" and strain, Experimental 

laws of . . . . . 750 

Stud bolt 1242 

Sun and planet motion . . .891 

Sun key 1246 

Supports, Reactions of . . . 771 
Surface of a beam. Neutral . . 789 



T 

Tap bolt .... 

Taper of cotters 

Tenacity .... 

Telodynamic transmission 

Tension .... 
" pulleys . 

Tensions in rope, Ratio of 
" in suspended rope 

Thermodynamics 

Thread of a screw 

Threads, Forms of 
" Gas pipe 

Tightening pulleys 

Toggle joint 

Tooth contact, Law of 

" outline, Epicycloidal 

Torsion and shafts . 

Train, Back gear 
" Engine lathe . 

Trains, Differential bevel 
" Epicyclic 
" Four-wheel . 
" Higher-wheel 
" Two- wheel . 



PAGE 
1242 

1259 

300 

1350 

749 

>349 

1350 

1351 

304 

347 

1224 

1231 

863 

824 

907 

908 

804 

875 

87s 

893 

888 

893 
S91 



Trains, Wheels in . . . 
Transmission of power by belts 
Trapezoidal screw threads 
Tredgold's approximation 
Triangle of forces 
Twisting moment 



PAGE 

. 872 

. 852 

. 1228 

. 918 

• 311 

. 826 



U PAGE 

752 

752 

301 

d screw thread 1224 



348, 81 



Ultimate elongation 

" strength 

Uniform velocity 
United States standar 
Unit of power 

" of work 

" strain . 

" stress . 
Universal joint 

" joint. Double 

Unstable equilibrium 

V 

Vapor .... 

Variable velocity 
Vein, Contracted 
Velocity 

" Formulas f(jr 

" of efflux 

" of flow, I\Iean 

" ratio 

" Uniform 

" Unit of . 

" Variable 
Vertical shear . 
Vibrating link motion 

^v 

Wall brackets .... 
Watt's parallel motion 

Wedge 

Weight 

" arm of a lever 

" Laws of . 
Wheel and axle .... 

" work .... 
Wheels in trains 
Whitworth quick-return motion 

" screw thread . 

Width of belt .... 
Wood in machine construction 

Work 

'' Unit of ... . 
Worm gearing .... 

"• gearing. Construction of 

" gearing. Friction of 

Wrenches 

Wrought iron .... 



353 
352 
749 
748 
835, 1287 
836 
334 



PAGE 

298 
301 
388 
301 
302 
384 
383 
2, 849 
301 
3c 2 
301 

773 
829 



PAGE 

1309 
897 

347 
299 

336 
318 
339 
340 
872 

833 
1230 

853 
1223 

352 

352 

920 

1326 

1327 
1235 

743 



m 



,\ W'M 



LIBRARY OF CONGRESS 





225 268 2 






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